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NOTE BOOK 


WIRING TABLES 


HOW THEY ARE MADE AND HOW 
TOAUSE= CHEM 








(SECOND EDITION) 





THE COOK-GRIER WIRING TABLE 
AND 


WIRING FOR MOTOR CIRCUITS 
BY 


THOMAS G. GRIER 





Copyright 1898 by Thomas G. Grier 





WIRING FOR MOTOR CIRCUITS—Copyrighted 1892. 
THE COOK-GRIER WIRING TABLE (Published in ‘‘Thos. G. Grier’s 
Note Book of Wiring Tables’—Copyrighted 1893. 


NOTE BOOK OF WIRING TABLES—How they are made and how 
to use them— Coyprighted 1897, 





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PREEAGE, 


IRING TABLES are a convenience and a 

\X) necessity to those engaged in Electrical 

Work. In preparing the second edition 
to this book the first edition has been carefully 
revised and new material added. 

The object has been to show how Wiring 
Tables are compiled, giving the fundamental 
reasons in as clear and simple manner possible, 
leading up in the methods used and in conclu- 
sion giving valuable information that will be of 
assistance in solving many of the questions that 


arise in practical work. 


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INDEX: 


THE RESISTANCE OF WIRES. 


Page 
The law of resistance;s ous. eos se OR ae i 4 
Explanation of the law 2. 20 a ae ny seed) os 7, 8,9 
The'‘definition of A: Civculay Mal. Tena cto) dee 9 
The relation of the'area to the diameter’ >= 2-2 2.) se 9 
The ohm (the unit of ‘resistance) >". =... }. +. = 0 9 10 
Table of comparative resistance of different metals .......... It 
Explanation of table. s..0.". 6. Sis GR s ee cena, £F, 32 


ELECTRO-MOTIVE FORCE AND CURRENT. 


i s\n fe) b de ee Ry EPA Gane co Gen OM Se ee A po) 
TPHEGA NI PETE Foe saree Fe NS, an a a ees eee an Le ee Re Art Gi 
Illustration of Volt and Anrpere®.. .-. 0.5. 4. «00 «7 ae pee 13, 14 
Ohims Taw 7. a he, ee Jae Set ad ets toes a ast ameligl spel ean 14 
Explanation of Ohms-law’: ... 4% ge: + hee eee Pe 14-95; 10 
The Watt, Electrical Energy or Power:..-.< < 26 up a 16, 17 
The: Horse) power 3°. 0). 4. Gye ects is on len ee tine Bote ee on 17 
The Candle powér 4) -. apeiron a eleanor ee ee 17 
The transformation of Electrical Energy... . .'.... wa ee 18 
DPAOLE-AULPEVES PEF A AULD ©. etic etree ie PEABO TETE A OS Gos Ge 18 
Explanation of table'amperes per lamp ...... . 95.) sue 19 
Explanation of table. amperes per motor... 24 cy-0ce eee a) er tobe site 20 
Table. Amperes per Motor’... 5 co sae a SO ; 21 


HOW TO CALCULATE THE SIZE OF WIRE. 


General description and explanation ........... oie ve the oe Mea TROL: 
General description condensed in six separate and consecutive state- 
ITLETUCS co Deec cs ve wie dee wel ane. fe, Matis. Yaeger oe ital oe (niet eeteneas nt 25 
Statements on page 25 repeated in different form ........... 26 
Simple formula™ sco 5 5 ites so on oto teen «tents elo) omer ee 26 
Table Resistance of Copper at various temperatures. ......... 2} 
Explanation of Current Carrying of Wire... . . 5:6, eee 28 
Table of current carrying capacity of wire... 2s 2). < 29 


4able of fusing point of 8metals .. . 1. . 5k es = 6 eee 30 


METHODS OF WIRING. 


Multiple arc wiring — Explanation and diagram. ........°.. 31 
Series System of wiring sf ao \ ra ce cares Nair reseaeeroe ob chars 32 
The series multiple system of wiring—Explanation and diagram .. . 32, 33 
The multiple series system of wiring—Explanation and diagram . . 33, 34 
Edison three wire system—Explanation and diagram ........ 34, 35 
Beedersia vemos. es 3 SHOE Phe eee AMEE RES SOIC Ut Mg aaa 35 
CNG C Com ME SRI ee Se a ah Ot. oh Sita gine ate Gulley. be Oy a foe) eldeeee 35 
SIeICORLG te OFC CISLLIDIILIOIN Ss) ue) i ee ce A elem Mich el one eee Shas 35 
Mee ometneOMReCGers CLG ci eas so Me oncile. sine ck Foca teh ohee elo cea ata 36 
Application of formula in calculating the size of wire for any of the 
Bere IOUS OL WiIlINe fai 's. 6 chee Sis ww A LbaY yee Bee ee tie a77 38, 80 
COMMERCIAI, WIRING TABLES. 
PE LCeT ere COMMIS FR DIC Le. cri ale clap wee vere sls gels cee AOTAT A 2 
Hxamples showing workingof tables .........6..... 42, 43, 44 
sua lemomone per CeNLIOSSI5O-VOILS| (5 0. she es i ee Se ae ee 41 
os ** two i BOF AGO ST “Sane cee ee ee ene ae es 42 
+ ““ two ie Ze Hig toy., ARES "ca oft ca Sf tee ane a ae oe a Bs 43 
& “* two ss Peo, gS oe 8 lie”, Ss, A ee ee mee Beli 44 
os «« 50 volts from 2 to 10 per cent loss, from 1totooamperes. . 45 
14 ae 100 ac ae 2 “ce Io ae oe “ce I ce I0o 46 ew | 46 
as ee 220 ae ace 2 ae Io ae ae ae I “ce 100 ae ey 47 
ac oe 500 ae ‘ec 2 ae IO ae sé ing I ce 25 “6 ian 48 
sé OO nee aaeT CO ed ss <3 Ole LOO “ 8 SAG 
os STOO ee ame aeeeT © oe alah, 25 of oe «50 
oe Buen OOO} © AS} OY 5 fo: ok ut Ses O mal OO s Pn Re 
es i 2000 ce ve 2a TC =e .s ae 4 STOO ce a es 52 
ee OE Xoyere) ke See 2 TO uN a see 2 coe TOO oh ein Be 
a HS = Troveley = siemens te “ an 2:5. LOO te se 54 
Table showing difference between wire gauges ............- 55 
Pepa GiMmenisions Of pure Copper Wite .-.-50.. 6 5 8 Ew a als 56 
Mabie: resistance of pure copper wire... 3 s.5 see pe ee Ee. 57 
Comparative table of diameter and weight of copper wire ...... 58 
Wetpnte.or iron, steel; copper and brass wire. ..%-...... re ee! 
Complete table of the Metricsystem of measurements. ........ 60 
ammeroiaeciita Wecnnivalennts (<2 %s.i.. +82 este es ee ee ee 61 


Diagrams and descriptions of special switches and their wiring con- 

MEO stot We" bN6 sls. cys sue seer te tal leka™ ote, evielsije ts. fertsc'e Oar 0S 
SRMPMMEtCION ittnie Lables « os 6 0 d 5 os w fe 8 #0 s 6 6 46m 4) 5 04-19 
Development of the simple formula. 5. - 3 5 6 1 8 ws 8 ee Aaa GibrY, 
Facts about the Brown & Sharpe Wire Gauge ....56..+5.2-+-24-+2e 0 
Bimeaotatation(Cook:-Gricr) «3's «0-0 © ere 6 % sc 6.0) Ses, 08 eULTIO 
SinEPL OGD Ix=(rtiGL weber eed cl ols 's o.ne) smrel es 167 6 us) 6. ‘of 4. ere te ep cers 75 
Peuanrrespey per Olt 5 6) 6) sl 2. oe 6 «86 ous |e 0.81 4 swivels oo 


wy = 
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Weight of weatherproof wire, table. ......-. 


Cost of copper, table... . 


oo. eG @-¢e «€ 8 @ 


\WTberb eke? Torr iaeKonope eRe 4 A A Go ba 


Mit peres PeLanOLOL table wsemcms nme 
Minimum size wire for motor service, table 


Wiring for motor service, table .....- 
Large size feeders with ground returns . 


Standard diagrams 


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. 80-88 
eR 

85 
er ae 
. 89-90 
. 91-99 


THE RESISTANCE OF WIRES. 


To determine the sizes of wires for the distribution of elec- 
tricity from one point to another there are three factors which 
enter.iuto the problem. 

The resistance (ohms). 

The electromotive force (volts). 

The current (amperes). 

In all substances there is something which offers an obstruc- 
tion to the flow or transmission of the electric current; this 
obstruction is technically called resistance. 

It is not known what resistance is, but the more practical 
problems as to the degree it exists in different substances, and 
as to how it varies in the same substance or material, have 
been solved. 

Resistance in various substances varies; that is, a piece of 
iron does not offer the same resistance as a piece of platinum 
of equal size, and copper offers less resistance than iron. 

In the same substances the resistance varies with the dimen- 
sious, and the law is stated as follows: 

“tHE RESISTANCE OF ANY SUBSTANCE VARIES 
DIRECTLY AS ITS LENGTH, AND INVERSELY AS. ITS 
AREA OF CROSS-SECTION.” 


Ne 
7 


x ‘ye 
\ ws 


Mri SoloLANCEK: VARIES “DIRECTLY “AS <THE 
LENGTH,”’ hence the resistance of two feet of wire is double 
that of one foot of the same wife, and ten feet of wire will 
have ten times the resistance of one foot; thus if the resistance 
of any length of wire is known, the resistance of any other 
length may be found. 


Example :—The resistance of 1,000 feet of number 10 B. & 
S. copper wire is one ohm, what is the resistance of 500 feet ? 

Answer:— 1,000 feet have one ohm’s resistance, one foot 
will have the one-thousandth of one ohm resistance, or .ooI 
ohms, and 500 feet will be 500 times the resistance of one foot, 
or 500 times .oor ohms, which is .5 ohms. 

Example:—-A certain wire one foot long has two ohms 
resistance, what is the resistance of ten feet of the same wire? 

Answer: — Ten feet is ten times as great as one foot; there- 
fore the resistance will be ten times as great, or ten times two 
ohms, or twenty ohms. 


* 
x * 


“THR RESISTANCE VARIES INVERSELY AS THE 
AREA OF CROSS SECTION.”’ To illustrate ‘‘the area of 
cross-section ’’ so that it will be clearly understood, take a bar 
two inches thick and three inches deep, cut this squarely in 
two, and there is an end, the dimensions of which are two 
inches by three inches ; this end represents the cross-section, 
and its area is two times three, or six square inches. The area 
of cross-section of a wire is the area of a circle, the diameter of 
which is the same as the diameter of the wire. 

“THE RESISTANCE VARIES INVERSELY?)=ihaeee 
when the cross-section increases, the resistance decreases ; or 
if the cross s-ction decreases, the resistance increases. 

If a wire of a certain area of cross-sec‘ion had one ohm resist- 
ance, a wire of one-half the area of cross-section and same 
length would have twice the resistance, or two ohms; or if a 
wire the same length had ten times the area of cross-section, 
the resistance would be one-tenth of one ohm, or .I ohm, 

Example :—An iron wire has ten ohms resistance, how much 
larger must a wire of the same length be to have one ohm 
resistance P 

Answer :— The resistance is to be decreased from ten to one; 
therefore the area must be increased from one to ten, or in- 
creased ten times. . 


Example :—A wire has a resistance of one ohm per thousand 
feet, how much smaller or larger must a wire of the same mia- 
terial be to have twenty-five ohms to the thousand feet? 

Answer :— As the resistance required of the same length is 
larger, the size of the wire must be decreased, and therefore the 
wire will be smaller; the resistance is to be increased twenty- 
five times, therefore the area of cross-section will be one twen- 
ty-fifth that of the first wire. 


x 

Wy 
» , 
% 3 
ee 


It is necessary before going further to define the units of 
measurements, wires are circular and their area of cross section 
is the area ofacircle. The unit of area isa CIRCULAR MIL. 
We are accustomed to measuring areas, when of small dimen- 
sions, in square inches; the areas of the cross-section of wires, 
however, are so comparatively small that the inch is divided 
into the thousandth part—-one thousandth of an inch being 
called a mil, and the areas being measured in square mils. As 
the area of cross-section of a wire is the area of a circle, it is 
awkward to calculate the square mils, and the unit circular 
mil has been adopted. A CIRCULAR MIL is the AREA of a 
circle whose diameter is a mil (one-thousandth of an inch) and 
is equal to .7854 square mils. 


* 
* * 


THE AREAS OF CIRCLES ARE TO EACH OTHER AS 
THE SQUARE OF THEIR DIAMETERS. This means that 
a circle of two mils in diameter has four times the area of a 
circle one mil in diameter; the square of the diameter is the 
diameter multiplied by itself. 

If the area of a wire one mil in diameter is one ci:cular mil 
- the area of wire three mils in diameter is equal to three multi- 
plied by three, or nine circular mils. The area of a wire one 
hundred mils in diameter is 100 times Ioo, Or 10,000 circular 
mils. 


Example : -—What is the area of a wire one inch in diameter? 

Answer :—One inch is 1,000 mils and the square of 1,000 is 
one thousand multiplied by itself, or 1,000,000 and the area is 
1,000,000 circular mils. 

Example :— One wire is 7 mils in diameter, another 14 mils 
in diameter, how much larger is the one than the other ? 

Answer :— One wire has 7x7, or 49 circular mils; the other 
has 14x14, or 196 circular mils and 49 into I196is 4, which shows 
that the area of cross section is four times larger in one than 
the other. 


The OHM is the name used for the unit of resistance. Dif- 
ferent substances offered different resistances, and mercury was 
selected as the most suitable material, upon which to deter- 
mine the standard. 

The legal ohm is the resistance of a column of mercury one 
square millimeter in cross-section and 106 centimeters in 
length, at a temperature of 32 degrees Fahrenheit. 


* 
aha 


A copper wire No. to B. & S. gauge 1,000 feet long, which 
has an area of 10,381 circular mils, offers a resistance of about 
one ohm. Mercury at 32 degrees Fahrenheit has about 58 
times as much resistance as copper. 


* 
Kk # 


DIFFERENT SUBSTANCES of the same size do not offer 
equal RESISTANCE. It has been found that when a wire of 
copper was used there was less resistance than there would be 
if the wire had been iron or some other metal. It was also 
found that copper offered the least resistance, and with it as 
a basis a table of comparison was made. 


10 


TABLE OF RESISTANCE. 
Comparative resistance of various metals with annealed copper. 


NAME OF METAL Ween Wie cea eee ue 

(PURE ) y in cet mil aS resist- 
Bewerannealed.. .. . . . .» .9.936 ohms, ~O2 
Sompermaunesledis: fo... 9.718 1.00 
Copperhard drawn ...7. +. 2° 9.9407 ‘“ 1.02 
Gold annealed. . et nd eel 2.520 pars: 1.28 
Aluminium annealed . Bee Tic 20) mek 1.80 
PertesseUaet cc. Ge 5 | 632.220. 3 20 
Platinum annealed... . wees 57000 5.67 
Propanmenledie saan 59.400 «= SS 6.11 
GIO Mee Me cid. a eee 75.760,» 7.80 
re mesceC rh a aes at 2 5 /00.300/, ** S527 
Meg orsesseCes a sc, . 119.390.) ~** L225 

German silver, hard or an- 

SIE CME Seng PU area? yp 275320), -/ 9S! 13.10 


The resistance in are above table is determined at freezing 
point, or 32 degrees Fahrenheit. 

The temperature has an influence on the resistance of sub- 
stances ; iron increasing in resistance rapidly as the tempera- 
ture increases ; other metals increasing in a greater or less 
degree. Hence, when comparisons are made, the temperature 
must be considered. In examples following the comparative 
resistance as given in the above table will be used. 

Example :— What size iron wire, whose length is one hun- 
dred feet, will be required to furnish an equal resistance to 
1,000 feet of a certain sized copper wire? 

Answer :— The resistance of iron is 6.11 times that of copper, 
therefore if one thousand feet of iron was to offer equal resist- 
ance to one thousand feet of copper, the iron wire would have 
to have 6.11 times the area of the copper. 

The resistance decreases with the length, and to have 100 feet 
of iron wire of equal resistance to 1,000 feet the size would 
have to be decreased by ten. Therefore the size of 100 feet of 
the iron wire to equal the resistance of the 1,000 feet of copper 
wire, would be one-tenth of 6.11, or .611 times the area of the 
copper wire, 


li 


ty 


Example :— What is the resistance of a wire 100,000 circular 
mils in area, and 12,000 feet long, if the resistance of a wire of 
the same material 10,000 circular milsin area and 9,900 feet long 
is 9 ohms? 

Answer : — The first wire is ten times as large in area as the 
second, and one and one-third as long. Its area is increased 
by ten, therefore its resistance would be decreased to one 
tenth; but the length is increased to four-thirds, therefore the 
resistance is equal to four-thirds multiplied by 9 ohms divided 
by ten, or one and two-tenths ohms (1.2 ohms). 

Example: — If in the above the first wire is iron and the sec- 
ond wire copper, what would be the resistance of the first wire? 

Answer :— The resistance of the first wire when both wires 
were of the same material was 1.2 ohms. Iron has 6.11 times 
the resistance of copper, therefore the resistance would be 1.2 
multiplied by 6.11, or 7.332 ohms. 


Electromotive Force and Current. 


(VOLTS AND AMPERES.) 


The electromotive force is that peculiar property of electri- 
city which overcomes resistance, or, rather acting against re- 
sistance produces results—The greater the electromotive force 
the greater the results. What electromotive force is is un- 
known, but how to produce it and how to measure its effect 
by comparison has been accomplished. To compare anything 
when there is a standard to compare it to, is to measure it. 

The VOLT is the name given to that standard and it is 
called the UNIT OF ELECTROMOTIVE FORCE or Electri- 
trical Pressure. (The expressious, difference of potential and 
voltage are also used, meaning electromotive force.) 

Electromotive force acting against resistance produces what 
is called electric current. This current is also measured by its 
effect, and the unit is called the Ampere. 

An AMPERHE is a current of such strength as would deposit 
from solution .006084 grains of copper per second. 

A VOLT is such an electromotive force as would cause a cur- 
rent of one ampere to flow against the resistance of one Ohm. 

To illustrate : —Supposing there is a circuit, including a de- 
vice for holding a solution of copper, the total resistance of 
which is one ohm, an unknown electromotive force is applied 
for a definite number of seconds ; after finding the amount of 
copper which has been deposited by the electric action and 
dividing it by the number of seconds this action had been 
going on, the amount deposited per second would be known. 
Should this amount be .006084 grains then the current strength 
would be one ampere and the electromotive force would be 
One volt. Laboratory instruments are callibrated in this 
manner. 3 

These units, the volt and ampere, are derived from what is 
called the C. G. S. system, an explanation of which is not 
necessary at this point, it being sufficient if the meanings of 
the terms are comprehended. 


13 


To measure electromotive force and electric current the 
magnetic effect of the current acting against either springs or 
gravity is utilized in the majority of measuring instruments in 
use, 


OHM’S LAW. 


As has been stated the greater the electromotive force the 
greater the results) THE CURRENT STRENGTH IS DI- 
RECTLY PROPORTIONAL TO THE VOLTS, AND IN- 
VERSELY PROPORTIONAL TO THE OHMS. This is 
OHM’S LAW and expresses the relation which the three units 
bear to each other. 


1st: THE AMPERES ARE EQUAL TO THE VOLTS 
DIVIDED BY THE OHMS. 

2d:—Also, THE VOLTS ARE EQUAL TO THE AM- 
PERES MULTIPLIED BY THE OHMS, and 

3d:—THE OHMS ARE EQUAL TO THE VOLTS DI- 
VIDED BY THE AMPERES. 


The above are three ways of stating Ohmis law. 
IF ANY TWO FACTORS ARE KNOWN THE THIRD 
CAN BE FOUND. 


Example: —If there is a circuit on which 100 volts is ap- 
plied and the resistance is known to be 10 ohms, what will be 
the current? Answer :—100 (volts) divided by 10 (ohis) will 
give 10, therefore the amperes are Io. 


Example :—If amperes are Io and the resistance 50, what 
are the number of volts? Answer :— 10 (amperes) multiplied 
by 50 (ohms) is equal to five hundred, therefore the nusnber of 
volts are 500. 

Example : —If the volts on a circuit are 110 and the amperes 
50, what are the ohms? Answer:—TIIo (volts) divided by 50 
(amperes) is equal to 2.2, therefore the ohms.are 2.2. 


14 


To assist in making the conception of electromotive force 
clearer a comparison is frequently given to the flow of water. 
Electromotive force is pressure, and we speak of so many 
volts pressure as we speak of so many pounds pressure or so 
many feet of head in a water system—the greater the head of 
water, or the greater the pressure, the more water flows 
through a system of pipes, and analogous to this, the more 
electrical pressure the more current flows. 

In a water system, the farther you are removed from the 
source of supply the less the pressure is. The pressure at the 
eud of a mile of pipe is less than at the pump; the pressure is 
lessened by overcoming the friction and transmitting the 
water, but the water is not lost. And so it isin asystem of 
electric wires, the pressure is lessened as the distance from the 
source increases, but the current is not lessened. The illus- 
tration carried farther, supposes the pump to have a pipe run- 
ning from its discharging end through a system of pipes and 
back to its suction end; the pipe is filled with water and the 
pump set in motion. If the pressure in the pump be small 
the water will circulate through the pipe slowly, using up all 
the pressure but the quantity of water remaining the same. 
The more we increase the pressure the greater will be the rate 
of flow of water. 

We may vary the resistance in the pipes by placing water- 
wheels in the system and have them do work. The pressure 
will be partly used up in making them turn, and part in the 
pipe. The more we increase the pressure the more water will 
pass the wheels and the more work they will do. 

In electric transmission the action is corresponding in effect 
to that of water. A wire leads out and another returns to the 
electric generator. The higher the pressure is raised, the 
greater the rate of flow of current ; the pressure being lost, the 
amperes current returning. 

The pressure decreases gradually as the distance from the 
generator increases and in proportion to the resistance over- 
come. If the total resistance is ten and the circuit simply a 


15 


loop of wire in which the resistance is uniform, then at one- 
tenth of the distance one-tenth of the pressure has been used, 
and so on until atthe return end it is found that all of the pres- 
sure is gone. If the generator continues to generate a con- 
stant pressure it will be found that, when measuring for the 
amperes they will be found the same at one-tenth, at two- 
tenths, or any part of the circuit. 

Example :—A generator produces a constant pressure of 50 
volts and has an iron wire circuit of Io ohms resistance, what 
is the current at any point? what pressure is lost or used at 
one-tenth ? at one-half? 

Auswer :—The current is equal to the volts divided by the 
resistance, or 50 divided by Io, which is 5, or 5 amperes. The 
total pressure is 50 volts; at one-tenth of the distance the re- 
sistance of one-tenth the total resistance is overcome, there- 
fore one-tenth of the total pressure is used, or one-tenth of 50, 
or 5 volts. And at one-half of the distance it will be 25 volts. 

It may be explained also: as one-tenth of the distance has 
one ohm resistance, the amperes are 5, the resistance 1. The 
volts are equal to the amperes multiplied by the ohms, there- 
fore the volts used in this portion of the circuit are 5 multi- 
plied by 1, or 5 volts; and at half the distance the resistance is 
5 ohms, which, multiplied by 5 amperes give 25 volts for one- 
half the distance. 


* 
ELECTRICAL ENERGY OR POWER—THE WATT. 


The production of electrical pressure and current requires 
work and the value of electricity depends upon the ability with 
which it can be changed into other forms ofenergy. By means 
of the incandescent lamp and arc lamp we transform electricity 
into light, by means of the motor we turn it into mechanical 
power—the telephone, the telegraph, the electric bell, the 
annunciators, etc., are means of utilizing this unknown force. 

In all cases it requires both electrical pressure and current 
combined, 


16 


The unit of electrical power is the WATT (the volt-ampere) 
it is the product of the volt and ampere. 

Example : — How many watts in a circuit in which the pres 
sure is 50 volts and the current 5 amperes. 

Answer: —5 multiplied by 50, or 250, which is the product 
of the number of volts multiplied by the number of amperes; 
which gives 250 watts as the result. 

The WATT, in mechanical units is equivalent to 44.25 foot 
pourids per minute or .7375 foot pounds per second. Arc 
lamps take from 300 watts to 600 watts to develop the nominal 
2,000 candle power. Motors require from 825 to 1,000 watts 
per horse power developed, depending upon the size and make 
of the motor —the difference between 746 watts (which is an 
electrical horse-power) and the amount required represents the 
loss in the transformation of electrical energy to mechanical 
energy. 

In the transmission of electrical energy for power or light- 
ing it is necessary first to determine the amount of mechanical 
power desired and determine from the ability of the motors 
used how much electrical power will develop the mechanical 
power (Zz. é., overcome the losses in the motors in addition to 
the power required. ) 

Also in regard to the lights it must be known how many 
candle power can be produced for the expenditure of so many 
watts. When these are known then the problem of finding the 
proper size wire for transmitting the current is a problem 
easily solved. 

The following tables have reference to the average efficiency 
of motors and lamps in commercial use: 

A HORSE POWER is a mechanical unit and is the work 
done is raising 550 pounds one foot high in one second or 
33,000 pounds one foot high in one minute. 

Seven hundred and forty six WATTS equal one HORSE 
POWER. A kilowatt is 1,000 watts. 

A CANDLE POWER is the unit of light. A candle which 
burns 120 grains of spermaceti per hour or 2 grains per minute 


17 


is supposed to give an illumination equal to one standard 
caudle. 


* 
% * 
THE TRANSFORMATION OF ELECTRICAL ENERGY. 


In producing electrical energy the generators develop volts 
and amperes ; or watts, the product of the first two. 

Motors and other devices produce mechanical power, lamps 
produce light. It must, therefore, be remembered that when 
generators are asked for to develop so many horse power or so 
many candle power, the efficiency of the motors or lamps must 
first be considered. Some incandescent lamps require as many 
as six watts to develop one candle power, while others take but 
three watts for each candle power, and better than this has been 
claimed. 














LAMPS. AMPERES PER LAMP.’ 

VOLTS 52 79 110 220 
IO AS ke. 6 43 eRe 22 
16-5 TO .68 a 33 
Sone 1325 .86 .64 .40 
2 a 154 1-10 .80 .50 
Le 2.00 1.36 I.00 .65 
SO 3.10 2:20 1.60 I.02 
TOO? 2% 6.20 4.30 3.20 2.05 
TSOne 9.30 6.40 4.75 S37 








The above refer to the lamps of these voltages. 

If ona 220-volt system two lamps are used of I10 volts each, 
the pair of lamps take the same amount of current as on the 
110-volt ; 7.é., two 16 C. P. lamps will take .5 amperes. On the 
500-volt railway system five r10-volt lamps are used in series— 
each set take approximately the amount set opposite the 
lamps; 2. é., five. 16 C. P. lamps at 500 volts would take .5 
amperes. 


18 


On the 1,000-volt system alternating when the secondary 
current is 52 volts, the transformation is about 1 to 20, and as I 
ampere represents a 16 C. P. lamp on the secondary, I-20 of an 
ampere at 1,000 volts represents the one lamp on the primary. 

On the 1,000 volts when the transformation is I to 10, a 16 
C. P. lamp is .5 amperes and I 20-ampere at 1,000 volts in this 
case represents about one 16 C.-P. lamp. 


This information enables us to change the candle power to amperes. 


Example :—How many amperes are required to supply seventy 16 C. P. 
lamps on a IIo-volt system. 


Answer :—One 16 C. P. lamp takes .5 amperes, therefore seventy lamps 
takes 70 times .5, or 35 amperes. 


Example :—How many lamps can be supplied with eighty-six amperes 
ona 75-volt system. 


Answer :—A IoC. P. lamp takes .43 amperes, this divided into eighty- 
six gives 200 10 C. P. lamps. 


Example :—Ou an Edison three-wire system 220 volts, forty 16 C. P. 
lamps are burning what are the amperes. 


Answer :—One 16 C. P. lampat iro volts takes .5 amperes, and two in a set 
to make 220 volts, only take .5amperes, therefore if two take .5 amperes, 
40 will take just 20 times the amount two will take, or 10 amperes. 


Example :—On a 220-volt system where one lamp is made for 220 volts, 
how many amperes wi!l forty lamps take. 


Answer :—One lamp takes .33 amperes and forty takes forty times .33, or 
about 13.2 amperes. 


Example :—When 100 16 ©. P. lamps are on a 1,000-volt circuit, the second- 
ary being 52 volts, how many amperes are on the secondary, and also on 
the primary. 


Answer :—One 16C. P. lamp at 52 volts takes one ampere, 100 lamps 1oo am- 
peres; on the primary, or 1,o00-volt circuit 1 20-ampere is equivalent to 
one 16C. P. lamps, therefore the current is 1-20 of 100 or 5 amperes. 


Noter.—In figuring for the size of wire if the voltage used 
happens to be a few volts above those marked in the table the 
same amount set opposite the size of lamp can be used. 


1 


AMPERES PER MOTOR. 


Whenever a certain horse-power is to be developed and the 
electromotive force is known this table will show at once the 
amperes required. 

This table is arranged to show the amperes per motor at the 
average commercial efficiencies indicated for various horse 
powers up to 150, and various voltages up to I,200. One col- 
umn shows the watts per motor, and another shows the number 
of 16-candle power lamps that equal the energy the motor 
draws from the circuit. 

Example :—What number of amperes are required to develop 150 H P. 
at 220 volts. 


Auswer :—565, follow the column marked H. P. down until 150 is reached 
and then across until the column 220 is reached, this gives 565 amperes, 


NoTrE.—If the power is to be developed by 150 motors of one 
horse-power each the result will be different, for the efficiency 
of asmall motor, is less than a large one—that is, it requires 
more electrical energy to develop a proportional mechanical 
energy. One horse-power motor at 220 volts takes 4.5 am- 
peres, and 150 will take 150 times 4.5 or 675 amperes. 

Thus it is seen that before the size of wire can be determined 
upon conditions must be known. 


























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21 


HOW TO CALCULATE THE SIZE OF WIRE. 


The size of a wire for transmitting electrical energy is calceu- 
lated upon conditions governing each individual case, yet the 
fundamental basis and method never changes. 

The resistance to be overcome in a circuit consists of two 
parts, first, that offered by the wire, and second, that offered by 
the apparatus or devices which convert or change the electrical 
energy into other forms, the two resistances use up the total 
energy, the first without any apparent returns, while the 
second produces available results. The problem which pre- 
sents itself is to proportion the wire so that the loss is consist- 
ent with the existing conditions and satisfies both the com- 
mercial and engineering features. 

Thus the calculation of the size of wire is simply the find- 
ing out what size will allow the transmission of the current with 
the loss that has been predetermined ; for instance, if the loss 
decided upon is two per cent, then the size to be calculated is 
that of a wire which will use up two per cent of the total en- 
ergy supplied, allowing 98 per cent available for power, light- 
ing or heating. 

x» 


As has been shown in the preceding chapters we cannot de- 
termine the size of wire without first reducing the number of 
lights or the horse-power to amperes and volts. Tables from 
commercial data were given to enable the student to obtain 
the amperes when the voltage, together with the number of 
lamps or horse power, is given. 

As a further assistance it may be repeated, that 746 watts is 
the equivalent of a horse power, and that the total) number of 
horse power, multiplied by 746 watts, equals the total number 
of watts. 


22 


However, in a motor the available horse power, which is 
frequently called the mechanical horse power, does not repre- 
sent the total horse power supplied to the motor, for a certain 
amount of electrical energy is lost or used up in overcoming 
friction, in heating and in magnetic losses. Thus in estimating 
the watts required by a motor we must add to the available 
norse power (which is usually the rated horse power of a motor) 
the amount required to overcome the losses spoken of in order 
to arrive at the correct number of horse power, which we mul- 
tiply by 746 to get the total number of watts. 

As the amperes multiplied by the volts equal the watts, 
then when we have obtained the watts we can divide by the 
volts and this will give us the amperes. 

In regard to incandescent lamps, the maker usually will give 
the watts or the amperes or the volts, and sometimes their re- 
sistance, and with such data you can easily calculate the 
“‘load’’ or the total number of amperes and the voltage of 


your system (the electromotive force). 


* 
% * 


According to ohms law it requires the pressure of one volt 
to produce or transmit one ampere over the resistance of one 
olm. It requires two volts to transmit one ampere over two 
ohms. As the resistance is increased the volts must also be 
increased to keep the current constant. 

It is also true, according to this same law, that two volts 
will transmit two amperes over one ohm and three volts will 
transmit three amperes over one ohm, thus it is seen that 
when the resistance remains constant the number of amperes 
increases as the pressure or volts increase. 

When the volts remain the same and the resistance in- 
creases, the number of amperes decrease, and when the resist- 
ance decreases, the number of amperes increase; that is, one 
volt pressure acting against one ohm can transmit one ampere, 
but when acting against two ohms it will be used up in trans- 
mitting one half an ampere, while if the resistance is but one 
half ohm, it will transmit two amperes. 


23 


The whole problem of finding the proper sized wire for 
transmitting current or electrical energy is the problem of 
proportioning resistance. 7 


We know that one volt acting against one ohm will transmit 
one ampere, and the volt be completely used up in doing the 
work; we therefore reason that for each ampere we wish to 
transmit we must have one volt for every ohm of resistance, 
and we would multiply the number of ohms by the number of 
amperes to see what number of volts would be required; but 
in the case of determining the size of the wire we wish to use, 
the resistance (the ohms) is unknown — but we have the 
number of volts and the number of amperes. We know that 
the total number of ohms in the circuit must be equal to the 
volts divided by the amperes. 


It is not, however, the total resistance of the circuit which 
we must have as the resistance of the wire; the volts divided 
by the amperes gives us this total resistance which is repre- 
sented by the resistance of the apparatus (lamps, motors, etc.) 
plus the resistance of the wire, the total volts are entirely used 
up in overcoming this total resistance. What we wish is to 
determine a size of wire that will take or use up a commercial 
or proper proportion of the total volts. This proportion is ar- 
bitrary, depending upon both engineering and commercial 
judgment. The theory is, first obtain your total estimated 
resistance by dividing the pressure (volts) by the total maxi- 
mum amperes and then use a wire, the resistance of which is 
one tenth, or one fiftieth, or one one-hundredth, of this total 
resistance, or any proportion of this total resistance which 
may be decided upon. 

* x 

In the foregoing no reference was made to any particular 
metal or any table of wire sizes. In the practical everyday 
transmission of electrical energy, copper is the metal em- 
ployed. Copper wire is made in standard sizes, the areas of 
which are given in circular mils. 


24 


The practical method for determining the size of copper 
wire is to obtain the results not in resistance but in circular 
mils. To do this we must remember the relation which resist- 
ance bears to the area and the length of wire, which is that 
the resistance increases directly as the length and inversely as 
the area; we must also decide upon the units of measurement, 
which, at the present time, are the foot for length and the cir- 
cular mil for area. As the resistance is the factor which gov- 
erns the size of the wire, we must know the resistance of our 
unit, which is a wire one foot long and one circular mil in 
area. This resistance, which has been obtained by experiment, 
we will assume as 10.6 ohms (refer to table of resistance of 
copper at various temperatures). 


I. The resistance of a copper wire is equal to 10.6 ohms 
multiplied by its length in feet, and this product divided by 
its area in circular mils. 

2. The total resistance of any circuit is equal to the volts 
divided by the amperes. 

3. The resistance of the wire is equal to the total resistance 
of the complete circuit divided by the proportion decided upon, 
or multiplied by the per cent, to be lost in the wire. 

4. If we wish to find the area of a copper wire and know 
the voltage, the number of amperes and the per cent loss, (or 
the proportion the resistance of the wire bears to the total 
resistance): 

We first consider that the resistance of this wire would equal 
(its length in feet multiplied by 10.6 ohms) divided by (its area 
in circular mils), and from this we reason that the area is there- 
fore equal to the (length multiplied by 10.6 ohms) divided by 
(its resistance). 

5. The resistance, however, is equal to the (per cent loss) 
multiplied by the (volts divided by the amperes). 

6. Therefore the area is equal to (10.6 ohins multiplied by 
the length) divided by (per cent loss) multiplied by the (volts 
divided by the amperes). 


25 


The above are rather lengthy statements and when given by 
formulas may seem clearer. While algebraic expressions are 
not always best in work of this kind, it is hoped that they will 
not be confusing when given, following the above statements. 


Let R represent the total resistance. 


ee a. 4 resistance of the wire. 
aw = at volts 

eA a ve amperes. 

ey i et distance in feet. 

Fyies per cent of loss. 


a - area in circular mils. 


this transposed, or multiply- 








statement (1) >). « pes ing both sides of the equation 
by @ gives (4). 
v 
4¢ ray 
(3) ers R=— 
iat 2 ee Ree eg ame 
(10.6 
; tAIA [fee 
rs 
¥ (5) gol g the value of R as given in (2) is 
pt Pees eee * inserted in (3) and gives for- 
mula (5) 
Ay (6) aren 6X/_10.6X/XC 
en he Top % Be = q E 
& 
IODC 


Now this formula—- is simply the number of am- 


% 
peres (C) multiplied by the TOTAL LENGTH OF WIRE multiplied 
by the constant—and the whole product divided by the volts 
lost, for the per cent of E (FE stands for the total volts) is 
the volts lost. 

Note, we say the total length of wire; we have been figuring 
on the resistance of the wire and have taken the length which 


26 


includes both the outgoing and return circuit. In the follow- 
ing chapter the method of figuring for the various kinds of 
wiring will be taken up and the above formula applied to fit 
the different conditions. 


RESISTANCE OF COPPER AT VARIOUS TEMPERATURES. 


In the table of comparative resistances given in a preceding 
chapter, it was stated the comparisons were made at 32 
degrees Fahrenheit. Substances at different temperatures do 
not have the same resistance ; some increase in resistance rap- 
idly as the temperature increases ; others change but little, 
while a few are known to decrease as the temperature in- 
creases (carbon). 

Copper, the metal which is the one considered in our wiring 
tables, has a variation of one and one-half ohms ina range of 
about 67 degrees Fahrenheit ; this is shown in the accompany- 
ing table: 


Resistance per Mil. Temperature in 

foot in legal ohms. Fahrenheit degrees. 
eh eS ee ere are awe Yo ae. a SOA 

TY LS Re eae a oe So 61 

We MES ata tea oat SEI si) hl! tee en ae 59.79 

Te eee gaol ea ke Se 14S 3 aS O4.40 

ea eR ee ng ee SLES care ts Ske 4 SOSLOF 

“ote. eS heaic be Oe aie a SP or cee Ac at 

SmI Race oe es ee fa a ek a ee, eS PBLOI 

ee Ee Pe aR gs kre. hp. Sd oe Beg OD AT 

Oo BR SE 7 aa A a Oo 28. 

TS/00"."'. MECN eerie ct cste ds ka. OLS lh 

Pe OMaP tes BGO vin ifs te xg een es ah & a 95-69 

Meet Ts gat yal ite ss as er aed 2) LO0.04 

Rarer a. i A Sard beg Gate anc, Ge) hy LOS ZO 

ae ee Alin wn ig a foo rst extn ert Pa wt pt OO.O4 

Se gh Peel in Og ans ae Peel wae pe. 112,90 

ESM Ge fea ce 6 gag We eles oe Oat wae a LEZ. TA 


The Current Carrying Capacity of Wire. 


The flow of current through a conductor produces heat; this 
heating effect is in proportion to the amount of current, and if 
the amount of current is large enough it will melt the wire. 
The same amount of current will not always heat the wire to 
the same temperature unless the existing conditions are the 
same ; as for example, a wire in winter exposed to a tempera- 
ture below freezing, and subjected to a current of Ioo amperes 
would be lower in temperature under this condition than it 
would be if the surrounding atmosphere were at a temperature 
of 80 or 90 degrees Fahrenheit. 


By the foregoing illustration it can be seen that the carrying 
capacity will depend within a limited range upon the sur 
rounding temperature, for the wire will allow an increasing of 
the current up to a point where its temperature becomes dan- 
gerous, either by cause of setting on fire materials with which 
it may come in contact, or, if nothing of a combustible naturs 
is near, its temperature will keep increasing as the current 
increases, until the wire melts. 


A wire will not become as warm when it is open and ex- 
posed to the circulation of air as when closed or confined. 
The radiation of the heat developed in the wire greatly affects 
its carrying capacity, and if the radiation is almost totally re- 
stricted a small amount of current applied continuously would 
melt the wire. 


Recognizing the danger in overloading wires in commercial 
use The Underwriters National Electric Association have 
established a table of safe carrying capacity, and the following 
is a quotation from their rules and requirements, ‘“‘Edition of 
Jan. I, 1896, page 22”’: 


“TABLE OF CAPACITY OF WIRES.”’ 


“Tt must be clearly understood that the sizes of the fuse 
depends upon the size of the smallest conductor it protects 


28 


and not upon the amount of current to be used upon the cir- 
cuit. Below is atable showing the SAFE CARRYING CA- 
PACITY of conductors (copper) of different sizes in Brown and 
Sharpe gauge, which must be followed in placing of interior 
conductors : 


TABLE A. TABLE B. 

CONCEALED WORK. OPEN WORK. 
Baus. G, Amperes, Belo. Gs Amperes. 
ae eid oe ie we 218 OOOO Fath ta te ee ol 2 
ooo . . 181 OOOSSIS Ere shes se202 
oO . . 150 oo . . 220 
oO . 125 Oo. eelos 
Ris na f€ i= cae 50 
2. 88 25 sees ye A 
ce 75 Ne ib ede en, ie IIo 
4. 63 4 92 
5: 23 See ia. 
6. 45 6. 65 
ae 23 3 46 
Io. 25 EOhey or, cae SU ee ee 
| Pine ok a a Oy, E2e, Wis tess ad es 
DAE Meten devas ke 2 TAS es. eee, 10 
boy Ce ee ee 16g ote re S 
RL ee, ca eee TONS iar ae eer bs sie ty 


‘““NoTE.—By ‘open work’ is meaut construction which 
admits of all parts of the surface of the insulating covering of 
the wire being surrounded by FREE air. The carrying capac- 
ity of 16 and 18 wire is given but no wire smaller than 14 is 
to be used, except as allowed under rules 27 (d) and 31 (a).” 


The pamphlet from which this quotation has been taken 
consists of thirty-nine pages and is entitled, ‘‘Rules and 
Requirements of the National Board of Fire Underwriters,”’ 
and may be obtained from any representative of the associa- 
tion or from W. H. Merrill, Jr., electrician for the association, 
at 157 La Salle street, Chicago. 


29 


The carrying capacity of wires depends also upon the uses 
aud conditions for which they are intended; the table just 
given is for copper wires and for their use in. distributing 
current from one point to another. 

Manufacturers of rheostats use both German silver and iron 
wire, surrounded or supported by non-combustible material. 
Each manufacturer, for his own use, has compiled from prac- 
tical experiments, tables applicable mainly for special work. 
These tables, while of use for other purposes, are not available, 
as they are of more value to individual manufacturers if kept 
secret. 

Some approximate results are given, however, showing the 
size of wires of different material which will be fused by 1oc 
amperes of current, when exposed, with a surrounding atmos- 
phere of between 70 to 80 degrees, the current being turned on 
for several seconds : 


Copper. 6.2 Set NON S 7 eb oa eae 
Aliminum io oie n ew aa ee we - “ 
Piatt sora ao Bare ee ee eS re = 
German Silver? st. eo aa ees 2s :: 
LE OUT sc eo cle yao ee ee eee — 
Ti ee eee ee Se Se ee Gime & s 
Lead o2.%.. 5 Sty hae ae ak Ce ee ee 3 3 
Tin and lead allows Sie hn dee, eect a aaa es 7 
- 


To give absolute accurate results for the fusing or melting of 
wires all of the conditions must be known, the temperature of 
the atmosphere, the medium surrounding the wire if it is con- 
fined, the length of time the current is on and the exact alloy 
of the metals and actual experiments, are the best method of 
determining the results. 

The figures just stated are given as a guide for comparison, 
similar to a table of breaking strains, such as are tabulated 
for various materials used in structural work, not so much 
with a view of using them as of giving data so that danger- 
ous ground may be avoided. 


30 


Methods of Wiring. 


MULTIPLE ARC WIRING is the system used most exten- 
sively for incandescent lighting and power purposes, and is 
frequently called wiring in parallel. The two wires run side 
by side, one the negative and one the positive, and lamps or 
motors are connected across from one side to the other, as 
shown in the diagram, A CONSTANT DIFFERENCE OF ELEC- 
TRICAL PRESSURE BEING MAINTAINED BETWEEN the two 
wires. 





THE MULTIPLE ARC SYSTEM OF WIRING. 


In this system the amount of current varies in proportion as’ 
the number of devices, for utilizing it, are increased or de- 
creased. 

The term coustant potential is applied to the system using 
this method of wiring. 

Among the devices used are constant potential arc lamps, 
incandescent lamps and motors. 

The present successful street railway system now univer- 
sally used is wired upon this method. Alternating stations 
have their line distribution upon this basis also. 

In regard to the alternating current, while one wire for a 
fraction of a minute is positive and the next instant negative 


31 


the DIFFERENCE IN PRESSURE is constant, the pressure simply 
changing in direction a number of times in a minute, causing 
the current to change in direction as frequently as the pres- 
sure changes, but not altering the difference in pressure be- 
tween the two wires. 


THE SERIES SYSTEM is in the nature of a loop, the 
greatest difference of electrical pressure being at the terminals 
of the loop. The current in this system of wiring is constant 
and the pressure varies, increasing or decreasing as the devices 
are cut in or cut out of circuit. The principle of this system is 
shown in diagram. 


t- +— +- 





=e 


THE SERIES SYSTEM. 


The devices used on this system are the same as those used 
in the multiple system but are constructed upon different 
lines. The series arc lamp, however, is the main device now 
operated commercially by this method, as many as 125 arc 
lamps being placed in series. 


THE SERIES MULTIPLE is a system where a number of 
multiple arc systems are placed in series; this method of wir- 


32 


ing was formerly employed in obtaining incandescent lights 
from an arc light plant, but is not approved of by the Fire Un- 
derwriters and is not used in this manner in any new installa- 
tions and has been taken out of nearly all places where for- 
merly used. 


THE SERIES MULTIPLE SYSTEM. 


THE MULTIPLE SERIES is a system where a number of 
small series are connected up in multiple. This method is 
employed on constant potential systems where the voltage is 
many times greater than the voltage required by the devices 
to be used. 

Our electric street car system presents a good example of 
this method—5o00 volts approximately is the pressure applied 
to street cars; the commercial incandescent lamp of the past 
five years was made for voltages no higher than I1o to 120 
volts, to light cars and car barns; 100 volt incandescent lamps, 
5 in number, were connected in series, and as each used up I0o 
volts, the total of the 5 would use up 500 volts, and each series 
would be placed between the 500 volts of the street car line. 
Storage batteries are connected up by any one of the four 


33 


methods described, in order to obtain different results, and 
many are the ways in which these methods are em »loyed 
singly or in combination. 


THE MULTIPLE SERIES SYSTEM. 


THE EDISON THREE-WIRE SYSTEM is in the nature 
of a multiple series. Incandescent lamps not being made to 
stand a higher pressure than slightly above IIo volts, a system 
was devised that could use from 220 to 250 volts. The first 
step was that of placing two I1o volt lamps in series between 
220 volts. However, when one lamp was turned off it would 
turn off the other lamp in series with it, and two lamps had to 
be turned off and on at the same time. ‘To avoid this a third 
wire was introduced, which was called a neutral wire ; this 
was placed between the two lamps of the series and run back 
to the generators, there being two generators or dynamos, 
each generating 110 volts, and the positive of one connected to 
the negative of the other. In this way, if there were ten 
lamps, eight lamps burning on one side of the neutral wire 
and two on the other side, the surplus current would flow back 
along the neutral wire and permit the turning on or off of 
lights at will. To balance this system requires either a large 
amount of copper in the distributing mains and feeders or 
special] arrangements at the central station. Incandescent 


34 


lamps, motors and constant potential arc lamps are operated 
on the Edison three-wire system. 

Heating apparatus and devices are also being introduced, 
and can be used on the multiple, multiple series and the Edi- 
son three-wire system with commercial success. 





THE EDISON THREE WIRE SYSTEM. 


FEEDERS is a name used to designate the wires which 
convey the current to any set of other wires and are a feature 
of the multiple, multiple series and Edison three-wire 
methods, . 

DISTRIBUTING MAINS are the wires from which the 
wires entering buildings receive their supply. 

SERVICE wires are the smaller wires entering the buildings 
supplying the various apparatus used. 


THE CENTER OF DISTRIBUTION is a term used differ- 
ently according to conditions—the first center of distribution 
is in reality the central station—this would be the center for 


35 


the whole plant. When simply mains are to be considered, 
the point at which the feeder is connected is the center of dis- 
tribution. 

Where houses or buildings are considered it may mean the 
point at which the service is connected to the mains, or where 
the service is connected to the wires of the building. 


= 
aes 


Feeders 






aes | 
Ula Ysssssitissttts 
Y Service 


S 


SAS’) : MMA 


Building Line 


VIEW PEEE SELIG 
Service 


oN 


DIAGRAM ILLUSTRATING FEEDERS, MAINS AND SERVICE. 

Then each floor of a building may have a cut-out box and 
these in their turn may be termed centers of distribution. 

When the distance of center of distribution is given to 
determine the size of wire for lamps or motors, it is well to 
ascertain whether it is the right center or not; perhaps it may 
be only the distance from the cut-out box that has been given, 
when it should have been the distance from the point at 


36 


which the service enters the uilding, or even greater, the dis- 
tance from the point at which the service is connected to the 
mains, for when the size is determined it is for a certain 
amount of current ; the transmission of additional current on 
the mains in the building increases the loss in volts in that 
main and likewise also in the service. 

Most buildings are wired upon the basis of a certain per 
cent loss in voltage, figured from the point at which the serv- 
ice enters; any additions to the wiring should be figured from 
the same point. 

The service is also figured from its point of connection at 
street mains to the point at which it attaches to the building 
wires. If additions are to be made to the interior wiring that 
will increase the number of amperes, then the service requires 
attention — otherwise the number of volts at the lamps or 
devices will be less than required and results will not be sat- 
isfactory. 


APPLICATION OF FORMULA. 


10.6x 7 xC . . e ‘ 
The formula, =~ equals the area in circular mils, is 


the basis for calculating the size of wires in all cases, and 
means simply that if you multiply the total length of wire by 
the maximum current, and multiply this by the constant 10.6, 
(the resistance in ohms of 1 foot of copper, one circular mil in 
area) and divide this by the volts lost, which is the per cent 
of the total voltage, you will have the area of a wire in circu- 
lar mils that fills the requirements. 

Apply this formulato a Multiple Arc System. The first point 
tc be determined is the length of wire, note the two wires are 
parallel, therefore the total length of wire is twice the total 
distance, and Z is twice the distance. You then ascertain how 
many amperes are on the circuit. This youmay know, or you 
may only have the load given in lamps and H. P.; in which 
case you reduce this to amperes. The voltage on the circuit is 
known in any particular case. You take the per cent of the 
voltage and divide it into the product of amperes multiplied 


37 


by the length as found, and this multiplied by 10.6, and the 
result is the area in circular mils. 

Example :——- What size wire is required for a 50 volt system 
having 10o lamps at a distance of 100 feet, having 4 per cent 
loss. 

Answer :— The load of 100 lamps on the 50 volt system is 100 
amperes—4 per cent loss is 4 per cent of 50 volts, or 2 volts. 
Applying the formula, we multiply the total length of wire 
which is twice the distance, or 200 feet, by the 100 amperes of 
current; this gives 20,000; we then multiply this by the con- 
stant which is 10.6, which gives a result of 212,oco—dividing 
this by the 4 per cent of 50 volts, or 2, and we have 106,000, 
which is the size of wire in circular mils that is required. 

In figuring the distance it is not always taken as the total 
distance, noris it correct to take the total distance, but the 
average distance, For illustration, suppose one light is 50 feet 
from the point from which the distance is determined, and the 
farthest lamp is 300 feet, and the lamps are distributed evenly 
between these two points, we would average up the distance 
between the first and the most remote lamp, which would 
be 150 feet, and add to this the 50 feet, making the average 200 
feet. You must use judgment in assuming this mean or aver- 
age distance, as the lamps or motors are buncked in different 
ways for each case. 

In a Series System the loss in voltage, while making as much 
difference in loss of power, does not affect the commercial 
phase of the question as much as in a multiple arc system, forin 
a multiple arc system the devices depend for their efficiency 
upon a constant pressure, and it is important that the loss in 
the conducting wires must not exceed the amount predeter- 
mined. However, ina series system, the apparatus depends 
upon a constant current, and the voltage varies with the resist- 
ance to keep the current constant, and this is done by a regu- 
lator upon the generator, which is designed to take care of 
changes of resistance in the circuit and increase or decrease 
the pressure as required. 


38 


In figuring the size of wire for a series system, for the length 
you take the total length of the loop, there is no mean or 
average distance, as the total current travels the entire dis- 
tance. 

If you have an arc light plant and are already using a 
No. 6 B. and S, gauge wire and want to find what the loss in 
the wire is—you have the area to start with and also know the 
amperes of your lamp and the length of the circuit, from this 
you can figure the loss. 

Example:—There are 10,000 feet of circuit, the lamps take 10 
amperes and the wire is No. 6 B. and S. wire, which has an area 
of 26,25@ circular mils—what is the loss in the line? 

Answer:—We multiply 10,000 feet by 10 amperes and this by 
10.6, which gives us 1,060,000, and divide this by 26,250 which 
equals about 40 volts lost in the wire itself. 

The Multiple Series, which is a number of small series con- 
nected in multiple, is virtually a multiple arc system, and the 
wire is figured from the formula as the multiple arc system. 

The Series Multiple consists of a number of small multiple 
arc systems but these are connected in series by the main wire 
and the wire is figured by the formula as shown for the series 
system. 

The Edison three wire system is a double multiple, the two 
outside wires are considered only when the wire is being 
figured, as when the system is under full load the neutral wire 
does not carry any current. Common practice makes the 
neutral wire one half the size of the other wires when the work 
is inside of buildings, though theoretically the neutral wire 
should be the same in size as the others. The neutral wire in 
feeders and mains depends upon the judgment of the engineer, 
and all three wires are frequently made the same size so they 
can be interchangeable, as a ground upon a neutral wire is of 
less consequence than on the negative and positive wire, and 
when any ground occurs within a section of mains the neutral 
can be changed and the grounded wire be made the neutral; 
this applies particularly to underground work. 


39 


Commercial Wiring Tables. 


Wiring tables were devised to arrive at results with thé least 
amount of work. 

To obtain exact results it would be necessary to fulfill all 
conditions of the formula, that is, the absolute constant should 
be used, the exact number of feet and the exact number of 
amperes. To compile tables upon this basis would require 
great care and they would be very large. Commercial require- 
ments do not necessitate such fineness. 

In compiling tables the number of amperes given are in 
such quantities that would seem to fill the nearest commercial 
requirements, and the same may also be said of the distances, 
the constant upon which the tables are based is one that 
would apply to average conditions, and the size of the wire is 
given in commercial sizes which are the nearest to the calcu- 
lated area, that will meet the requirements, which means that 
it is the commercial size which is larger (not smaller) than the 
area obtained from the calculation. 

In solving practical problems by tables one must take the 
figures for amperes and distance that will be the nearest to the 
actual conditions. 

Two tables are given for 50 volts, one for I per cent loss or 
4 volt loss, the other for 2 per cent or I volt loss; these tables 
are compiled upon the basis of 10.6 ohms for the constant and 
for multipleare wiring. The distance does not mean the length 
of the circuit (that is, the total length of both the positive and 
negative wire) but the direct distance between the lamps and 
the center of distribution. 


40 


y TABLE A. 
WIRING TABLE FOR ONE PER CENT LOSS ON 50 VOLTS (TABLE 
FOR 14 VOLT LOSS). 
















































































wn 
ro) 
= 
a Distance in Feet to Center of Distribution, 
o (Wire sizes are indicated in B. and S. Gauge.) 
ee 
° 
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7, |20/25) 30| 35] 40] 45 50| 60; 70} S80] go} roo} 120) 140] 160} 180} 200] 250 
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pe rOMmnG 9LS| sald jasTAte 13) Tale *r2|)) opr! TTs Ol PLO 9 8 8 7 6 
3/16/15] 14] 13 Be tele tele tT, ee TOO 9 9 8 a Tha Kein Ke 5 
Aisi a|s 12)" rit | IT)" —10}, “ro 9 § 8 7 7 6 5 5 4 gi 
Sirs lele) etd) «13/5 10 9 9 8 Gi 7 6 6 5 Ale a a 2 
6pl3|t2) 11) TO| -1o| .-9 9 8 7) 7 6 6 5 4 4 3 3 2 
7|I2|II| ro] Io} 9g 8; 8 7 7 6} 35 5 Ae 3 2 2 I 
8/11/10] 10 9 8 8 7 7 6 5 5 4 4 3 2 2 I fe) 
g|II|10] 9 8 8 7 a, 6 5 5 4 4 a 2 2 I I fo) 
1o|10} 9] 9 8 7 7 6 6 5 4 4 3 3 2 I I oO} 00 
12/10] g} 8 | 7 Bie 36 5 4 4 3 2 2 I I fe) o} 00 
14] 9] 8] 7 7 Giese 25 Site ie 4 Ali =3 2 2 1 fe) 0} 00} 00 000 
Tole) OF 6 5 5 4 4 3 2 2 I T 0} 00] 00} 000 0000 
18| 8} 7] 6 G2 ae | keer 7 ed ee) 2 I I 0] 00] 00] 000} 000 0090 
20)/71 0] FO 5 4 4 3 3 2 I I fo) ©} 0} 000] 000}0000 0000 
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———— SS Te ecic8i220 Sa — 


In the table for one per centloss on 50 volts (Table A) the 
loss is % volt. This means that % volt will be lost in trans- 
mitting the number of amperes indicated in the vertical col- 
umn on the left to the distances indicated in the top row, when 
the size of wire is used that is on the same line with the num- 
ber of amperes under consideration and under the number of 
feet which the amperes are to be transmitted, for example :— 

What size wire will be required to transmit 20 amperes 60 
feet, when the voltage is 59 and the loss one per cent. 


41 


Answer :—No. 3, B. & S. Gauge. Follow the column marked 
amperes down to 20, then look along the same line until you 
come to the column headed by 60, and 3 is the number which 
indicates that No. 3, B. & S. Gauge wire is the size required. 

Table B, which is for two per cent loss, is the same in principle 
as Table A, and is one that approximates actual conditions, as 
secondary wiring on alternating systems is usually installed 
upon the basis of two percent loss, When the secondary 
voltage is between 50 to 55 volts this Table (B) can be used. 


TABLE B. 


WIRING TABLE FOR TWO PER CENT LOSS ON 50 VOLTS (TABLE 
FOR I VOLT LOSS). 









































v3 
* 
a. Distance in feet to center of distribution. 
5 (Wire sizes are indicated in B. and S. Gauge ) 
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42, 


Table B could be used for other voltages and have quite a 
range of application by using the following suggestions : 

Note.—While Table Bis called a TABLE FOR Two PER 
“ENT LOSs ON 50 VOLTS, it means that one volt is lost. This 
‘able could equally as well be called a ONE PER CENT Loss 
» 100 VOLTS, for the 1 per ceut of 100 and 2 per cent of 59 
are the same. 









































TABLE C. 
WIRING TABLE 2 PER CENT. LOSS IIO VOLTS. 

f 

“ 

a Distance in feet to center of distribution. 

= (Wire sizes in B. and S. Gauge.) 

es 

° 

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It simply represents the size wire that will offer enoug!! 
resistance to use up one volt in transmitting the amperes tO 
the distances indicated inthe table. Nowif it is remembered 
that two volts will transmit.twice as many amperes the same 
distance over the same wire, or transmit the same number of 
amiperes twice as far, and for three volts, three titnes as far, 
etc., we can have any problem presented within reasonable 
range and solve it by thistable. We will take several exam- 
ples, to illustrate : 

Ist. What size wire will be required to transmit 30 amperes 
500 feet at 2 percent loss 500 volts? 
43 


Answer! 2 per cent of 560 volts is 1to volts. A wire that 
will allow 30 amperes to be transmitted 500 feet with ro volts 
loss would use up I volt every 50 feet. Referring to the table 
we find 30 amperes will be transmitted 50 feet with one volt 
loss on a number 5 wire, which would make a total of 10 volts 


loss in 500 feet. 
TABLE D. 


WIRING TABLE 2 PER CENT LOSS 220 VOLTS. 








' 
’ 


>Tunber of } 
Amperes 


Distance in feet to center of distribution 
(Wire sizes in B. & S. Gauge ) 














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Example: 9 amperes are to be transmitted 1,000 feet on a 
100 volt system with 5 per cent loss. Give the size of wire 
cequired. 


Answer: Five per cent of 100 volts is 5, and five volts lost in 
1,000 feet means one volt for each 200 feet. Referring to table, 
9 amperes are transmitted over 200 feet on a number 4 wire, 
with one volt loss, which would make a total of 5 volts lost in 
7.000 feet. 

As a matter of convenience two tables, one for I1o volts 
aud one for 220 volts will be given, upon the same basis as 
Table B. The constant being the same (i. e. 10.6) and the loss 


being 2 per cent. 
46 








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"SLIOA OOO'S “‘SLINDOYIO YOH ATEVL ONIYIM 











Table Showing the Difference 
Between Wire Gauges. 


Birmingham. Brown & 
No. London, Stubs’. Sharpe’s 
DOOD ers os niet CAAT iid aselsver ems’ a AOAT ae cateioonne 460 
ONMiae osc ote AO sual Fes esjnere ie esas ADEM ofoolateerrearse 40964 
1) oe B80 weer eee teen Aste) RCO ODN enti 35480 
QO ccsereeee BA) te wees eee DOA (yates) wisest alates 32486 
dope tr fel: © BON we ee eee eeeee Peal ete iuye (sietele! curs: 28930 
Be aeieas eae OE pend sida tem siapaiaye OBA Prone taleen ints 25763 
aerate encase DEG fe petdialece © ote Siem DAO udm es siecle 22942 
4 938 ween wee cece DS eae Ree enlan aoe 20431 
Bosse seceee 2). | na DOr disinie aterm cuales 18194 
iewslgse se Pid boneocrpoescre Geta oe peo acraeae 16202 
7 Ree SAcn Shagace Zk Neg Gckiaiot a8 14428 
eiere ciiets ss 165 GE ees co-chasniets Graces 12849 
OY ceNioe Sag TAR oeg ee Misteetile/ste RUA Sieincist cvercietesie'e 11443 
A amass ee See he She Danas ors BSAA ete aikier ei. cacats 10189 
11 120 120 .--e-eeeeeee J 09074 
hat Seger EA LOD me ne oes es ae 109 . 08080 
heir es Senceseae OOS ate ects erecta OREM as are ne eutiess 07196 
Ih, Matinee O83 a tiaatiose anes LOSS veriaeternsit sisiels 06408 
a Larrea eee (TORS Tanne aeel orc (Oc2ue 05706 
A Giccsractuleks Yee OGD im feats sie ers oe MOGI. Dtaieeaa se ace 05082 
17 Pie: ee en St ee ee. POA Bare ca tees 04525 
Thee Wen Bae OAD erie Saiteate Od Oeenanctce tae 0430 
We els eee ag YT ous ates thee VAD E ee cos ctaratee be 03589 
OTe Seber sctecar O35 sea ccctas ait oie TBS eters Minthehosn e 03196 
Pate et sais GTS wide es ee eee HS Dipe erate tas ater ateoieh Te 02846¢ 
Drea at oe SB c O2O5 Diy see whe Foes s RQ OR as snieictensare ae 025347 
DA toate OO tee (0)-7 (CODEC APATITE oti te SacmrOee 022571 
PAINS SENET OME Aa gets Meneses Oy ere O22 eeu). stsnctas 0201 
ae Nae Ste ae 023 oe 2 Oametobt=accettorer. 0179 
26 .0205 Pa OU Steetaae tals sats 01594 
DA Liat Ae ae REE G1S7otmas aceesdae cs 1 Geteracttoe fares 014195 
OSES ers ative OIG eer nies. are cs AN ois aaseonee 012641 
Orel ss d'els's OD Stare. Sas nests Ol ome rors eae 011257 
SOR weara see scie 01375 4 AU Wb eerreriar de FOG 010025 
ei leeeAanes wares O12 Scan teehee Ol Oigmesicutetcs avers 008928 
74 eg ap APOE OTD Beta ees ose oiete OOO: tse e aacioe 00795 
Dae eases wins O10 2atreanctte se wits .008 -; -00708 
7h lertete eens Panes OO9S Sista cieceraes OO Tiaras ses 0063 
Tae es «15 O09 Beano asses OOD ais! to.cyee coneres 00561 
SOMA eel cto SOOT Sees teers Ravina yO OL teres actker,s ot 005 
D nbstawattecrs EUG: Wer ects Marc nif Uk ater ie teeter es .00445 
DOG aes ese QOD TD Beta ons Rovctehsl ae Sttcicts tarielteere .003965 
DOM Meee) tp cele uameceies Catron» dete stare .003531 
CES iia Hee 0045 ale ites ete EM the slevaye aletats of -003144 
55 


Table of Dimensions of Pure 
Copper Wire.* 








*] mile pure copper wire ___ 


1-16 in. diam. 




















REVISED. 
Weight and Length. 
~ 
; soe! Sp. Gr. 8.9. 
CE Diam. 
; F P it Lbs. Lbs. Feet 
4 | Mils. yee oa per per per 
‘ 1000ft.| Mile. }| Pound 
009 460.000)/211690.0 |166190.2 |640.73/3383.04 1.56 
000 |409.640/1678U5.0 |131793.7 |508. 12/2682.85 1.97 
00 |/364.800]1383979 0 |104520.0 |4U2.97/2127.66 2.48 
O |324.950}105592.5 | 82932.2 |319.74/1688.20 3.13 
1 /289.300| 83694.5 | 65733.5 /253.43/1338.10 3.95 
2 (257.630) 66373.2 | 52129.4 |200.98/1061.17 4.98 
3 |229.420) 52633.5 | 413838.3 159.38} 841.50 6.28 
4 |204.310| 41742.6 | 32784.5 126.40) 667.38 qook 
5 |181.940; 33102.2 | 25998.4 100.23] 529.23 9.98 
6 | 162.020] 26250.5 | 20617.1 79.49) 419.69 12.58 
7 144.280] 20816.7 | 163419.4 63.03) 332.82 15.86 
§ | 128.490] 16509.7 | 12966.7 49.99) 263.96 20.00 
9 | 114.430} 18094.2 | 10284.2 39.64] 209.35 25.22 
10 {101.890} 103881.6 | 8153.67 | 31.44) 165.98 31.81 
41 | 90.742} 8234.11] 6467.06 | 24.93) 131.65 40.11 
12 | 80.808] 6529.94) 5128.69 19.77} 104.40 50.58 
13 | 71.961] 5178.39} 4067.09 | 15.68; 82.792 |- 63.78 
14 | 64.084] 4106.76) 3225.44 12 44| 65.658 | - 80.42 
15 | 57.068} 3256.76) 2557.85 9.86) 52.069} 101.40 
16 | 50.820) 2582.67| 2028.43 7.82)" 41.292 | 12787 
17 | 45.257; 2048.20} 1608.65 6.20) 32.746 |. 161.24 
18 | 40.303) 1624.33} 1275.75 4.92} 25.970 | 203.31 
19 | 35.890) 1288.09} 1011.66 3.90} 20.594 | 256.39 
20 | 31.961} 1021.44) 802.24 3.09} 16.381 | 323.32 
21 | 28.462) 810.09] 636.24 2.45] 12.952.| 407.67 
22 | 25.347) 642.47) 504.60 1.95} 10.272} 514.03 
23 | 22.571) 509.45) 400.12 1.54} 8.1450} 648.25 
24 | 20.100) 404.01 317.31 1.22} 6.4593] 817.43 
2p) | 17.900) = (3201 44 251.65 .97| 5.1227] 1030.71 
26 | 15.940} 254.08} 199.56 .11| 4.0623) 1299.77 
27 | 14.195) 201.50 158.26 61] -3.2215| 1638.97 
28 | 12.641 159.80] 125.50 -48} 2.5548} 2066.71 
29 | 11.257 126.72 99.526 .38]} 2.0260} 2606.13 
30 | 10.025} 100.50 78.933 .30] 1.6068} 3286.04 
31 | 8.928 79.71 62.603 24 1.2744] 4143.18 
32 | 7.950 63.20 49.639 ra ke) 1.0105} 5225.26 
33 | 7.080 50.138 39. 369 15 8014} 6588.33 
34 | 6.304 39.74 31.212 12 -6354| 8310.17 
Bo 5 0-014 Br D2 24.753 10 -5039) 10478. 46 
36 | 5.000 25.00) 19.635 .08 -0997 | 13209 .98 
37 | 4.453 19.83 15.574 -06 -8170/16654.70 
33 | 3.965 152 12.347 05 -2513/21006.60 
39 | 3.531 12.47 9. 7923 -04 .1993/26487 .84 
rat 3.144 9.88 7.7635 -03 -1580)33410.05 
i 


13.59 ohms at 15.5° C. or 


59.9° Ff. 


1 circular milis .7854 square mil, 


56 


Table of Resistances of Pure 














Copper wire.* 
REVISED. 
No. Resistance at 75° F. 
Bs 
& : he 
Ss 2: es Ohms F oct Ohms. 
* | 1000 feet. | Per mile. Ohms. per pound. 
0000 -04904 -25891)20392.9 -00007653 
000 -06184 -32649}] 16172.1 - 00012169 
00 .07797 -41 168) 12825 .4 -00019438 
0 -09827 -51885] 10176. 4 - 00030734 
1 - 12398 -65460} 8066.0 - 00048920 
2 - 15633 -82543] 6396.7 -00077784|  : 
3 - 19714 1.04090} 5072.5 -0012370 
4 . 24858 1.31248) 4122.9 - 0019666 
5 31346 1.65507) 3190.2 -0031273 
6 -39528 2.08706} 2529.9 -0049728 
7 49815 2.63184! 90 6.2 - 0079078 
8 .62849 3.31843} 1591.1 - 0125719 
9 - 79242 4.18400} 1262.0 - 0199853 
10 .99948 5.27726) 1000.5 - 0317046 
11 1.2602 6.65357} 793.56 - 0505413 
12 1.5890 8.39001' 629.32 - 080364 1 
13 2.0037 10.5798 499.06 - 127788 
14 2.5266 13.3405 395 79 - 203180 
15 3.1869 16.8223 Blak -323079 
16 4.0176 21.2130 248.90 -613737 
17 5.0660 26.7485 197.39 - 816839 
18 6.3880 33.7285 156.54 1.298764 
19 8.0555 42.5329 124.14 2.065312 
20} 10.1584 53.6362 98.44 0.284374 
21; 12.8088 67.6302 78.07 5.221775 
22} 16.1504 85.2743 61.92 8.301819 
23| 20.3674 | 107.540 49.10 13.20312 
24) 25.6830 | 135.606 88.94 20.99405 
25) 32.3833 | 170.984 30.88 33 .37780 
26| 40.8377 | 215.623 24.49 53.07946 
27| 51.4952 | 271.895 19.42 84.39915 
28| 64.9344 | 342.854 15.4) 134.2005 





29] 81.8827 | 432.341 12.21 213.3973 





30} 103.245 | 545.133 9.686 | 339.2673 
31) 130.176 | 687.327 7.682 | 539.3404 
32| 164.174 | 866.837 6.091 | 857.8198 
33} 207.600 1092.96 4.831 | 1363.786 
34} 261.099 |1378.60 3.830 | 2169.776 
35] 329.225 1738.31 3.037 | 3449.770 
36) 415.047 2191.45 2.409 | 5482.766 
37| 523.278 |2762.91 1.911 | 8715.030 
38) 660.011 [5484.86 1.515 |13864.51 

39) 832.228 [4394.16 1.202 |22043.92 

40|1049.718 [5542.51 -9525 38971. 11 





1-16 in, diam, 57 


59.9° F. 








H,; 





=x ¢ S2o 
Cah datacat a ee cme 
Bae SAB 
Sri g/fui 2 
a5 Alesa 
4.5 = 
800 12s, 
666 1.50 
500 2.00 
363 Foci bs. 
313 3.20 
250 4.00 
200 5.00 
14 6.9 
125 8.0 
105 9.5 
87 1 I ee 
69 14.5 
“50 | 20.0 
31 | 32:0 
"92-1 45.0. 
‘14 | 70.0 
“11 | 90.0 


se weer to rcoce 


es oo ee ey 


se eee 


eee eeete er cooe 


se eeerlececcee 


ies i ey 


es ie ee ey 


eoeccot} se ceece 


es i ee oY 


Ce ce a | 


Ce eee ae ee 


6) MLS ore |) e) Asie a 


Ce eeca{ce vo sto= 





*] mile pure copper wire __. 13.59 ohms at 15.5° C, or 


—— og 


Comparative Table of Diameter 
and beh ae of migoee: Wire. 











Diam- 
eter 1n 


| Mils. 


| No. of 





S| Gauge 


460.0 
409.6 
364.8 


Noe 
‘is 


BCONooF WNHC © 
iS] 
oS 
r= 
Sy) 


| 324.9 
2389.3 
257 .6 
229.4 








yr ounds 


CM=d2\1 000 feet. 


























623 925 
516.76 
437.107 
349.928 
272.435 
244.15 
202.965 
171.465 
146.51 
124.742 
98.076 
82.41 
66.305 
54.354 
43.59 
3.964 
Z71:319 
20.853 
15 692 
12.789 
10.18 
7.268 
5.340 


3.708 
3.099 
2.373 
1.892 
1.465 
1211 
-9807 
7749 
-5933 
516 
4359 
3027 
-2452 
1937 
1483 


-07568 


AMERICAN GAUGE. BIRMINGHAM GAUGE. 
Mw ols c | 
Area in| eee ° Eig Area i in 
CM=d2 1.000 tee 52a 
211600 | 639.33 || 4-0 | 454 | 206116 
16785 | 507.01 || 3-0) 425 | 180625 
133079 | 402.u9 || 2-0 | 380 144400 
| 0) 340 115600 
105592 | 319 04 1| 300 90000 
83694 | 252.838 2| 284 80656 
66373 | 200.04 3 | 259 67081 
§2634 | 159.03 4] 238 56644 | 
5 | 220 48400 
41742 | 126.12 6 | 203 41209 
331lU2 | 100.01 7 | 180 32400 
26244 79.32 8} 165 27225 
20822 62.90 9] 148 21904 
16512 49 88 10} 134 17956 
13110 39.56 14), 120 14400 
10381 31.37 12; 109 11881 
8226 24.88 13 | 095 9025 
6528 19.73 14 | 083 6889 
5184 15.65 15 072 5184 
4110 12.41 16 | 065 4225 
3260 9 84 17 | 058 3364 | 

2581 7.81 18 | 049 2401 
2044 6.19 19 | 042 1764 

1624 4.91 
1253 3.78 20 | 035 1225 
1024 3.09 21 | 032 1024 
820 2.45 22 | 028 784 
626 1.94 23 | 025 625 
510 1.54 24 | 022 484 
404 1.22 25} 020 400 
320 97 26 | 018 324 
254 cid 27) O16 256 
201 61 28; 014 196 
PAG, .48 29 | 013 169 
127 38 30} 012 144 
100 20 31 | 010 100 
79 2k 32 | 009 81 
63 19 33 | 008 64 
49 15 34 | 007 49 

36 12 

28 10 
25 08 2 35)" 005 25 
18 .06 ; 
16 .05 | 36) 004 16 


EEE 


04843 


Weights of Iron, Steel, Copper 
and Brass wire. 


DIAMETERS DETERMINED BY AMERICAN GAUGE. 


| 













































































i - Sinaor Weight of Wire per 1000 Lineal Feet. 
oe laa Seagal (MUA aos 
Ze each No.|/\ es Steel. | Copper. | Brass. 
INCH. LBS. LBS. LBS. LBS. 
0000 | = .46000 560 74 f66.03 640.51 605.18 
000 | .40964 444.68 448.88 507 .95 479.91 
00 . 36480 352.66 355.99 402.83 380.67 
0 32486 279.67 282.3) 319.45 301.82 
J . 28930 22119 || 223-89 ogee 239.35 
2; .25768 TORSO el gideaD 200.91 189.82 
3 22942 139.48 140.80 159.32 150.52 
4 . 20431 110.62 111.66 126.35 119.38 
5 . 18194 87.720 83.548 | 100.20 94.666 
6 | .16202 69.565 70.221 79.462 75.075 
Gj . 14428 55.165 55.685 63.013 59.545 
8 | .12849 43.751 44.164 49.976 47.219 
9] .11443 34.699 35.026 39.636 37.437 
10 .10189 27.512 215182 81.426 29.687 
11 .090742 21.820 22.026 24.924 23.549 
12]; .08°808 17.304 17.468 19.766 18.676 
13 .071961 13.722 13.851 15.674 14.809 
14 . 064084 10.886 10.989 12.425 11.746 
15 | .057068 8.631 8.712 9.859 9.315 
16 .050820 6.845 6.909 7.819 7.587 
7 .045257 5.427 5.478 6.199 5.857 
18 | .040303 4.304 4.344 4.916 4,645 
19 | .035890 3.413 3.445 3.899 | 3.684 
20 | .031961 2.708 2.734 3.094 2.920 
21 028462 2.147 2.167 Disp. 2.317 
22) .025347 1.703 1.719 1.945 1.838 
23 | .022571 1.350 1.363 1.542 1.457 
24 .020100 1.071 1.081 1.223 1.155 
25 | .017900 0.8491 0.8571 .9699 0.9163 
.015940 0.6734 0 6797 .7692 0.7267 
.014195 0.5340 0.5391 6099 0.4763 
012641 0.4235 0.4275 .4837 0.4570 
011257 0.3358 0.2389 . 3835 0. 8624 
.010025 0.2663 0). 2688 3042 0.2874 
008928 0.2113 0.2132 .2413 0.2280 
007950 0.1675 0.1691 -1913 0.1208 
.007080 0.1328 0.1341 1517 0.1434 
. 006304 0.1053 0.1063 . 1204 0.11387 
-005614 .08366 .08445 -0956 0.0915 
. 005000 06625 .06687 .0757 .0715 
004453 05255 .05304 .06003 . 05671 
003965 04166 .04205 04758 . 04496 
.0038531 03305 .93336 03755 03566 
.003144 .02620 .02644 .02992 -02827 
ecific Grav.. 7.7747 7.847 8.880 4.16 
Wet per cu. ft.|| 485.874 | 490.45 554.988 | 528.386 


Ene 


59 


The Metric System. 
Metric Denominations and Equivalents in Denomina- 
Values, tions in use. 
WEIGHTS. 
Weight of what quantity 
Name. No. Grams, of water at max. density. 
Millier or tonneau 1,°00,000 = 1 cubic meter. 


Quintal = 10,000 = 1 hectoliter., 
Myriagram = 10,000 = 10 liters. 

Kilogram or Kilo = 1,000 = 1 liter. 

Hectogram = 106 = 1 deciliter. 
Dekagram = 10 = 10 cubie centimeters. 
Gram = 1=} 1 cubic centimeter. 
Decigram = -l= .1 cubic centimeter. 
Centigram = .01 = 10 cubie millimeters. 


Milligram .001= ILcubiec millimeter. 


Name. No. Grams. Avordupois Weight. 
Millier or tonneau = 1,000,000 = 2,204.6 pounds. 


Quintal = 100,000= 220.46 pounds. 
Mvriagram = 10,000= 22.046 pounds. 
Kilogram or Kilo = 1,000 = 2.2046 pounds. 
Hectogram = 100 = 3.5274 ounces. 
Dekagram = 10 = 0.3527 ounce. 
Gram = l= 15.432) crams: 
Decigram — joes 1.5432 grains. 
Centigram = Ok 0.1543 grain. 
Milligram = UO) bee 0.0154 grain. 


MEASURES OF LENGTH. 


Myriameter = 10,009 meters = (6.2137 miles. 
Kilometer = 1,000meters = Moet m. or 3,280 ft. 
Qin. 
Hectometer = 100meters = 828 feet and 1 inch. 
Dekameter = 10 meters =393.7_ inches. | 
Meter — 1 meter ——" S0.on OeMnes. 
Dec*meter = .lofameter= 3.937 inches. 
Ceutimeter = .0lof a meter= 0.3937 inch. 
Millimeter = .00lof a meter= 0.0394 inch. 
MEASURES OF SURFACE. 
Hectare = 10,000 square meters = 2.471 acres. 
Are =+  100square meters = 119.6 square yards. 
Centare — Isquare meter = 1.550square inches 


MEASURES OF CAPACITY. 
Name. No. Liters. Cubic Measure. Dry Measure. 
Kiloliter =1,000= lcu. meter = 1.308 cubic yards. 


0.388 fluid oz. 
0.27 fluid oz. 


.01 = 10 c. centimet. 
001 = le. centimet, 


69 


Centiliter 
Milliliter 


Hectoliter = 100= .1lcu.meter = 2bush. 3.35 pks. 
Decaliter = 10—10c. decimet. = 9.08 quarts. 
Liter aa 1— ‘Vexdecimet: — 0:003s'guart- 
Deciliter = .1=—.ic. decimet. = 6.1022 cubic in. 
Centiliter = .0l1—10c.centim = 0.6102 cubic in. 
Milliliter = .001= le.centim. 0.061 cubic in. 
Name. No. Liters. Cubic Measure, Wine Measure. 
Kiloliter = 1,000=— lcubic meter = 264.17 galls. 
Hectoliter = i100 = .lcubic meter = 26.417 galls. 
Decaliter = 10—10c.decimeters= 2.6417 galls. 
Liter = 1= lc. decimeter = _ 1.0567 qts. 
Deciliter = .1=.lc.decimeter = 0.845 gill. 


Table of Decimal Equivalents 


OF 


ths, 16ths, 32ds and 64ths of an Inch. 


FOR USE IN CONNECTION WITH. THE 


Sths. 
{= qely 
t= .250 
= eRe 
t= .000 
Som .625 
8=—=.750 
f= .870 

16ths. 


b= 0625 
$= .1875 


i= 3125 

p87 
— 5625 

156—= -6875 


8 8125 


a Ce hy thong 
1¢== 93 i) 


32ds. 
— aoe 
3 = .0937! 
as eee 
oo == .21875 


9 = 28125 
Ml B4375 
13. 40625 

— 46875 
1 53125 
19 __ 59375 


2 65625 
B= 71875 
3 78125 
27 = 84375 
9 90625 


3! 96875 


GAths. 


— .015625 
See 
> = .078125 
u==-109875 
9 = 140625 
== .171875 
13 203125 
15 = .284875 
bi== 265625 

61 


MICROMETER CALIPERS. 


19 296875 
21 328125 
= 39010 
2 — 390625 
7i= 421875 
29 453125 
ey bcs 
j= 515625 
3) — 546875 
31 578125 
== .609875 
41__ 640625 
4 == 671875 
45 703125 
==. 734875 
“0 — 765625 


_gl==.796875 


83 828125 
d= 859375 
51 — 890625 
= 921875 
Wye OOL ZO 
63 984375 


Special Switches and Diagrams for Wiring. _ 


A three-point switch is designed to be so connected that- 
lights may be controlled from two places. When the lights ° 
are on, the turning of either switch puts them out, and an- 
other turn of either switch relights them. ‘These switches are 
sometimes called three way switches and lazyman switches. 
The switches are made by a number of manufacturers. 


/7ains 
Lights 










IPL Switch — SPL. Switch 





Wiamve Fon 
3 Wree CSSwrrer, 









DIAGRAM FOR WIRING A 3-POINT SWITCH. 


» 


Two diagrams are given for wiring them both practically 
the same, one diagram showing a round switch, the other a 
square or oblong switch, which represents the push switch, — 


62 


‘ce 


When it is desired to control lights from more than two 
places, it is necessary to use two three-point switches and as 
_many more four-point switches as there are places from which 
it is desired to control the light. There can be as many 
additional four-point switches used as desired. 


IPL Switch 4PC Switch 3PL Switch 














Wraive For 
4+WIRE _Swrrcr. 


DIAGRAM FOR WIRING A 4-POINT SWITCH. 





Two diagrams are given for wiring the four-point switches, 
both diagrams being practically the. same, but the different 
form of switches being shown in each. 


62 


INTRODUCTORY 


TOSTHS 


COOK-GRIER WIRING GABE 


This table, or system of calculation, the Cook- 
Grier Wiring Table, is intended for rapid and 
approximate figuring of wires. It was first pub- 
lished in 1893, but not as complete as in this edition. 

The development of the simple formula which 
precedes the table is given, not as a portion of the 
table, but because the Cook-Grier table is based 
upon the simple formula and the peculiar construc- 
tion of the Brown & Sharpe Gauge as explained 
on page 70. 

A close examination of the table will prove well 
worth the trouble, as it is easily remembered when 
understood, and consequently of much assistance 
when calculations are necessary and tables and data 
not at hand. 


The Development of the Simple Formula. 


In following the development of a formula it is best to first 
state it and then follow with the explanation. 

To determine the size of a wire for any purpose the follow- 
ing formula for multiple arc or two-wire distribution is the one 
most commonly used : 


ert ts ‘ 21.33% Cx 1) 
Area in circular mils = eee 

21.2= constant for 2 feet of copper wire one circular mil in 
area. (2 feet being used, as it takes into consideration 1 foot 
of positive and 1 foot of negative wire—2 feet of wire, but only 
a distribution to the distance of 1 foot.) 


C=current in amperes. 
D=distance in feet to lamps. 
E=volts lost in conductor. 


In regard to E the problem issometimes and in fact most fre- 
quently given as so many per cent loss, meaning so many per 
cent of the total voltage—that is, 2 per cent loss on a 50-volt 
circuit is one volt, while 2 per cent loss on 220 volts is 4.4 volts. 

Amperes are never lost ; pressure or voltage only decreases, 


05 


THE EXPLANATION AND THEORY APPLYING TO THE DEVELOP- 
MENT OF THE FORMULA. 

The theory of the wiring table is simply one of proportioning 
resistances. The first point to be considered is, how is the 
resistance varied in any substance? ‘The law states that the 
resistance of any substance varies directly as its length and 
inversely as its cross section. This law has been verified by 
experiment and is universally accepted as a law of nature. 

The resistance varies directly as the length of any given’sub- 
stance, means that ifthe length of a wire is doubled, the resist- 
ance is doubled, and three times the length means three times 
as much resistance. 

The resistance varies inversely as the area of cross section, 
means that if the area of a wire is doubled, that is, if there is 
just twice as much metal for a given length of wire, the resist- 
ance is decreased one-half. If the area is increased three 
times, the resistance is decreased to one-third. In a wireof a 
given substance if the length is doubled, and at the same time 
the area is doubled, the resistance remains the same, or if the 
length is increased to three times as great, and at the same 
time the area is increased three times, the resistance remains 
the same. ; 

From this application of the law already referred to, it can be 
seen howit is possible to vary the size of a wire so that the 
resistance remains constant for any length, or how the resist- 
ance may be changed to suit any condition or circumstance. 

The next consideration is the action of the electric current. 
The pressure is measured in volts, and the rate of flow is meas- 
ured in amperes. The volts are all used up in forcing the 
amperes against the resistance. 

The resistance, if it were all in the wires, would mean a total 
loss of the electrical energy, because the volts would be used in 
doing work from which no results were obtained. It is neces- 
sary, therefore, to make the resistance of the wire which con- 
ducts the current but a proportion of the total resistance. The 


66 


rest of the resistance of a circuit being that which exists in the 
lamps and the other devices or apparatus which convert the 
electrical energy into commercial forms, as light, power or 
heat. 

To so compile a table and calculate the sizes it is necessary 
to cousider the relation which exists between the unit of resist- 
ance, the ohm ; the unit of pressure, the volt ; and unit of cur- 
rent, the ampere. 

One ohm resistance requires the pressure of one volt to 
transmit one ampere. Two ohms require two volts to trans- 
mit one ampere. One volt will transmit but one-half an 
ampere over two ohms, aud one hundred volts will transinit 
ten amperes over ten ohms. 

This is Ohm’s law, which, when given in the form usually 
found in text books, is as follows: The rate of flow, or cur- 
reut, measured in amperes, is equal to the volts divided by the 
ohms. The volts are equal to the amperes multiplied by the 
ohms, and the ohms are equal to the volts divided by the cur- 
rent flow or amperes. 

Knowing the relation of resistance to the flow of currents, 
and the pressure, the next point to be considered is the sub- 
stance used for the wires and ascertain the amount of its re- 
sistance for a given length and size, to be used asa unit for 
making the calculation necessary in compiling the table. 

Copper, commercially and physically, represents the best 
conductor for the transmission of the electric current. A dol- 
lar’s worth of copper will conduct a greater amount of electric 
current than any other metal of equal value, with the same 
loss of energy In the illustrations following, copper will be 
the substance considered, as the wiring tables used for ascer- 
taining the size of wire are for copper wire. 

The units by which wire is measured are the circular mil 
and the foot. The circular mil is the area of a circle whose 
diameter is one-thousandth of an inch. The foot is a unit 
familiar to all. 


The unit of wire, then, would be a wire one foot long, the 
area of which is one circular mil, and the resistance of this 
unit of copper wire is between ten ohms and eleven ohms, de- 
pending upon the temperature of the wire. 

Ten and six-teuths (10.6) ohms may be assumed as the re- 
sistance which approximates closely to the average at ordinary 
temperatures. Taking this as the basis for compiling a wiring 
table, it is known that to send an ampere through this unit of 
wire would require 10.6 volts. For, as stated in Ohm’s law, 
the volts must equal the ohms multiplied by the amperes. 
But in everyday problems, the known quantities are the volts 
and amperes, and the calculation is to find wires of proper 
resistance to suit the conditions. The resistance, or the ohms, 
must equal the volts divided by the amperes. To solve this 
problem, the volts are divided by the amperes, and we have 
the resistance. 

The resistance varies as the length and inversely as the area. 
It is the area of a wire which is desired, and one the resistance 
of which will equal the resistance found by dividing the volts 
by the amperes. The resistance of one foot of copper wire, 
one circular milin area, equals 10.6 ohms. The resistance of 
any wire is equal to its length multiplied by 10.6 and this 
divided by the area. Now, it is known that the volts divided 
by the amperes equals the resistance, and the volts divided by 
the amperes are therefore equal to the length of the wire mul- 
tiplied by 10.6 and divided by the area. 

In any problem of wiring we have the number of volts and 
amperes, also the distance the current is to be transmitted, 
and the only thing remaining to be found is the area. To find 
this, we multiply 10.6 by twice the distance or the total length 
of wire (in multiple arc work there is one outgoing wire and 
the return circuit, which makes the total length of wire double 
the distance), and multiply this by the amperes, and divide 
the product by the volts used, which gives the area of the 
wire. 


68 


The volts used are not the total volts of the system, but the 
volts lost in the wire. If the problem is given as so many per 
cent loss, it would mean a per cent of the total number of 
volts. To illustrate: If the wire is to use up 2 per cent of the 
pressure, and the voltage of the system was 50, then you would 
divide by 2 per cent of 50 or one volt, or, if the pressure was 
100 volts, you would divide by 2 per cent of 100 or 2. 

In compiling a table, let it be assumed for a 50 volt system 
fora regular increase of distances of ten feet, the following 
method is pursued: First, the per cent loss is ascertained, let 
it be 2 percent. This then would be one volt. Then one am- 
pere will be considered, first, as the total current flow, and the 
following calculations would be made: 10.6 would be multi- 
plied by to feet and by 2to get the length of wire, and this 
multiplied by 1 ampere and the product divided by volts lost, 
or in this case, by one which would give us an area of 212 cir- 
cular mils. This would be the size of a wire necessary to allow 
one ampere to be transmitted Io feet with one volt loss (or 2 
per cent of 50 volts). 

The next calculation would be for 20 feet distance, and as 
the resistance of the wire must remain the same, the area first 
found, namely, 212 circular mils, must be doubled as the 
distance has been doubled. For 30 feet to keep the resistance 
coustant, the area is increased three times, and so on, until the 
limit of the table, say 200 feet, is reached, when the area will 
be increased 20 times, as the length has been increased by that 
amount. 

Then the same method will be continued for two amperes, 
namely, multiplying 10.6 by the distance and by 2 to get the 
length of wire, and this multiplied by amperes and the whole 
divided by the volts lost. This is continued until the area for 
all the wires for each of the distances and for the various num- 
ber of amperes have been determined. 

As it is necessary to multiply 10.6 by the distance, multi- 
plied by 2, in every instance, 2 X 10.6 or 21.2 is used instead, 


69 


and the straight distance used. The whole resolves itself into 
the simple formula: 

21.2 X distance X amperes 

= area of wire in circular mils. 

Divided by volts lost 

When these areas are found, the commercial sizes of wire 
are inserted in the table which are nearest to the size found, 
but always larger, never smaller ; that is, if the area found in 
the calculation came between a No. o wire and a No. 00 
wire, the No. 00 wire would be the size used in the table. 


Facts about the Brown and Sharpe Wire Gauge. 


In compiling the size of wires in the now universally used 
Brown and Sharpe Gauge, the size of No. ovoo compared 
closely to the No. 0000 wire in the old Birmingham gauge, but 
the other sizes were smaller. 

The sizes in the Brown and Sharpe or American gauge, how- 
ever, bear a definite relation to each other which allows com- 
parisons to be made between them. 

The size No. 0000 is approximately twice as large as No. o 
and No. o is twice as large as No. 3, and so on down the list tak- 
ing every third size, which means that No. 3 is also twice as 
large as No. 6. 

No. ooo is twice as large as No. 1 and No. oo is twice as large 
as No. 2. 

If one is asked to get a wire twice the size of No. 1o he would 
ask for No. 7, or if asked for a wire half the size of No.5 he 
would count three down the list and get a No. 8 wire. 

This peculiarity of the Brown and Sharpe gauge has made 
the table of rapid figuring known as the Cook Grier wiring 
table of value and with this knowledge and the application of 
the simple formula it is readily understood. 


70 


The Cook-Grier Wiring Table. 


The table which has been called the Cook-Grier Wiring 
Table has been known by that name since the latter part of 
1888. 

It was first developed by C. S. Cook, to apply to 50 and 1,000 
volt circuits in the earliest days of alternating current work, 
before wiring tables were published. 

At a later date the writer compiled and adapted the table 
for all voltages. 

As a table for rapid and approximate calculation, it is of 





much assistance. The merits can best be appreciated by a 


careful investigation. 

In the early days of incandescent work wiremen became 
accustom< d to figure on a basis of 100 volts. The introduction 
of the Edison three-wire system, using 220 volts, and the alter- 
nating system, using 50 volts on the secondary and 1,000 volts 
on the primary circuits, made matters somewhat complex. 


As men became accustomed to changing from one system to 
another, they begau to devise means of making one table do 
for all systems. This was accomplished with sufficient accu- 
racy for their purpose by using the 100 volt table. When they 
had 220 volts to work with, a wire one-quarter the size indi- 
cated was taken ; if the system was to be 50 volts a wire four 
times the size indicated was the size necessary. 

The more advanced and ambitious wiremen, in addition to 
the tables, frequently would figure their wires from a simple 
formula, getting the size of the wire in circular mils; with 
the aid of a circular mil wire table they could then check up 
the commercial size of wire and also verify their regular wiring 
tables. 

To place the workman beyond the necessity of or absolute 
dependence upon a table except one that could be carried in 
his head instead of his pocket, the approximate method (Cook- 
Grier Table) for figuring wire at different percen ae of loss 
and different voltages was compiled. 

The basis for the table is the formula known as ‘‘a simple 
formula,’? which is published in many catalogues and text 
books. 

The development of the simple formula has been explained 


in another chapter. 


APPLICATION OF SIMPLE FORMULA IN COMPILING ‘‘COOK- 
GRIBRGeeLA BI 


The simple formula which gives the area of wire equal to 
21.2 multiplied by the amperes multiplied by the distance 
and the product divided by the volts lost, or 


12, 


Pie : at.2x%CXD 
area in circular mils equals ee rene 
when used for a definite problem, is reduced to a still more 
simple form. 

In compiling this table the basis was considered as 50 volts, 

with one per cent loss. This would transform the formula to 
aie : 202K CXD) 
Area in circular mils equals Se pacar 
the one-half being the volts lost, which is one per cent of 50. 

When the denominator of a fraction is one half, it is equiva- 
lent to multiplying the numerator by two (2) and this makes 
the formula, area in circuJar mils equals 42.4xCXD. 

This means that the area is 42.4 times the product of the 
current and the distance, or if we divide the area by 42.4 we 

are 
Sie ite equals CX D. 

A table was compiled from this by dividing the area of each 
‘wire by 42.4 and making a list of constants, which are supposed 
to be the value of the sizes of the wire, and we say that the 
current multiplied by the distance for 50 volt table with one 
per cent loss, equals the constant. This is practically the 
Cook-Grier table. . 

To apply it to all voltages and any per cent loss, the follow- 
ing points are to be considered. 

When the voltage is increased say to roo, and the per cent is 
still one per cent, it means that twice as many volts are to be 
lost in the wire, and in consequence, the constant found by 
multiplying the current and distance is divided by two, be- 
cause the resistance can be doubled as the force (that is, the 
volts to be lost) is increased twofold. 

If the voltage is 500 and the per cent loss is Io, then the 
number of volts lost will be 50 and the constant found by mul- 
tiplying the current and distance will be divided by 100, be- 
cause the volts lost are 100 times as great as one per cent of 


50. 


get the formula, 


RULE FOR WORKING TABLE. 


The explanations of this table and the developments leading 
up to it have been given in order to make the theory clear. 

By simply following the three rules given, the table may be 
used without reference to the explanation. 

RULES:—1. Multiply the number of amperes by the dis- 
tauce in feet to the center of distribution and divide by the 
desired loss, the loss being in per cent. 

2. Divide the voltage used in the problem, by 50. 

3. Divide the result of the first by the result of the second, 
and the size whose constant is nearest to the final result thus 
obtained is the desired result 

N. B.—In practical work never use the size smaller than the 
result thus obtained, when the result comes between the two 
sizes, but use the larger. Iu the table of constants the con- 
stants are not exact, but are made in even numbers to facilitate 
the work. 


EXPLANATION OF TABLE. 
The table is for atwo-wire multiple arc system... When the 
three-wire Edison system is to be used, the size of the negative 
and positive wires is determined as if it were a simple two- 


wire system, the neutral wire being made one-half the size of 
the negative or positive wire. 


THE SIZES OF WIRE ARE EASILY FOUND, AS THE 
TABLE IS VERY EASY TO COMMIT. TO MEMORY. 
THREE SIZES ONLY NEED BE REMEMBERED, AND 
THE OTHERS CAN BE DEDUCTED FROM THEM. 


THE CONSTANT OF NO. 3 WIRE I5 1300. 
THE CONSTANT OF NO. 4 WIRE IS 1000. 
THE CONSTANT OF NO. 2 WIRE IS 1600. 


74 





















i 


Gauge.) 





211,600. 
167,805 


133,079 
105,592 
$3,695 


66,373 
52,634 
41,743 


33,102 
26,251 
20,871 


16,510 


13,094 
10, 382 


8,234 
6,530 
5b 


4 107 
22257) 
2,583 


2,048 
1,624 
42253 
1,024 
810 
642 


Circular mils 
sectional area. 


(Brown & Sharpe 





Gauge No. 


Size. 


OO000 
000 


0O 


alah, 


Ow oO SION &wWhr 


— 


THE COOK-GRIER TABLE. 

















Brown & Sharpe. 











Constant. 


5200 
4000 


3200 
2600 
2. 00 


1600 
I 300 
1000 


S00 
650 | 
500 


400 


325 
250 





200 | 
102 
125 


TOO 





60 


50 
40 
30 


25 
- 20 


Approximate 
weight in 
p unds per 
1000 feet. 





650 
500 


400 
325 
250 


200 | 
162 5 
125 


100 
81.5 
62.5 








7 


5 





Actual weight 


per 1000 feet. 







pounds 













640 
508 






402 
320 
253 


201 


159.38 
126.40 






100.23 


79-49 
63.03 


49.99 
39-65 
31.44 


24.93 
19.77 
15.68 






12.44 
g.86 
Fae 









6.20 
4.92 
3.90 
3-09 


2.45 
1.95 






To remember these three gives a complete clew to all the 
rest of the constants. If but one of these three is remem- 
bered, the fact that THEY DIFFER BY JUST 800 will aid the 
memory in holding them. 


To find the others is a simple mental calculation, EVERY 
THIRD WIRE GOING UP the table INCREASES TWO- 
FOLD; that is, the constant for No. 10 is 250, and going up 
the table three numbers we have No. 7, which is 500, or 
double No. 10. This is true of all sizes. (See facts about 
Brown & Sharpe table.) 


TO FIND THE APPROXIMATE NUMBER OF POUNDS 
PER t000 FEET, THE CONSTANT IS DIVIDED BY 8, 
aud gives a result that is very close and most convenient for 


making estimates. 


To use the table for determining the size of a series circuit, 
the total length of the circuit divided by 2 should be used for 
th- distance to center of distribution. 


Mr. Cook has further developed the ane of the table 
for rapid mental estimates, 


When it is desired to know the number of FEET PER 
OHM, multiply the constant by 4. The accompanying table 
has been compiled, giving the size of wire, the constant, the 
actual number of feet per ohm and the approximate result ob- 
tained by multiplying the constant by 4. 


The weight of bare copper and the weight of insulated 
weather proof wire do not have any definite relation, and can 
not be memorized by any easy method; a table has been com- 
piled, showing the weight of bare copper, the weight of triple 
braided insulation, the weight of the insulation, and the per 
cent of the insulation, using the weight of bare copper as the 
base to figure from. 


TABLE OF FEET PER OHM. 





Piraten ac sharye Constant. Feet per Ohm. Constant x 4. 


Gauge ) 





20, 392.9 20,800 
16, 172-1 16,000 
12,825.4 12,800 


10,176.4 10,400 
8,066.0 8,000 
6,396.7 6, 400 


5,072 5 5 200 
4,022.9 4,000 
Re1Q0s2 3,200 


2,529 9 2,690 
2,036.2 2,000 
1,591.1 1,600 


1,262.0 T, 309 
1,000 5 1,000 
793.56 800 


629.32 640 
499.06 500 
395-79 400 
313.87 320 
248.40 240 
197.39 200 


156.54 160 
124.14 124 
98. 44 100 


78.07 So 
61.92 60 
49.10 50 











WEIGHT IN POUNDS 


TRIPLIE BRAIDED INSULATED WEATHER PROOF WIRE. 














Weight of 


Approximate 
weight of 


Percent in- 


Weight of the| Ctease itt 





























Size. bare cop- [triple braided), nculation per| Weight tak- 
per per weather proof] { ooo feet. ing bare 
Brown & Shar 1,000. feet. wire per 1,000 copper as 
Gauge ) feet. 100 — 
0000 640.73 739 | 98.27 15% 
000 508. 12 598 97.88 19% 
oo 402.97 485 $2.03 20% 
Oo 319 74 395 75-26 23% 
I 253-43 313 59-57 23% 
2 200.98 251 50.02 25% 
k 3 159.38 205 45.62 28% 
4 126.40 168 41.60 33% 
5 100. 23 139 38.77 38% 
6 79-49 EPS y coh se oye 42% 
7 63.03 Sh. ra eo ee es m fouke 
8 49-99 74 24-01 50% 
9 39 65 ee ad re és eee 
10 31.44 5I 19.56 62% 
Il 24 93 Pash ae ig te 
12 19.77 37 1723 87% 
13 15.68 eee ben ae oe 
14 12.44 23 10.56 86% 
15 9.86 ne eee 2 awe plane 
16 7.82 16 8 16 104% 
17 6 20 toe 
18 4.92 12 7.08 1.33% 
19 3.90 <i wee ot ae 
20 3.09 





To quickly estimate the cost of bare copper wire, multiply 
the constant by two and point off two places for cents; this 
gives an approximate cost of copper at 16 cents per pound. 
The accompanying table shows the actual cost at 16 cents per 
pouud and the cost estimated by multiplying the constant by 
two. 


THE ESTIMATED AND ACTUAL, COST OF WIRE 


AT 16 CENTS PER POUND. 


























Size. ‘ Cost per 1,000 feet Constant x 2 and 
ee & Sharpe} Constant. at 16 cents per Ib, two places for cenis 
rauge.) 
0000 > 5200 $102.40 $104.00 
000 4000 81.28 80.00 
oO 3200 64.32 64.00 
O 2600 51.20 52.00 
I 2000 40.48 40.00 
2 1600 32.16 32.00 
3 1300 25.50 26.00 
4 1000 20.22 20.00 ; 
5 800 - 16.04 16.00 i 
i 
6 650 [2°72 13.00 j 
7 500 10.98 10.00 
8 400 8.00 8.00 i 
9 325 6.35 6.50 ! 
10 250 5.03 5.00 
! 11 200 3.98 4.00 
12 160 3.16 3.0 
¥4 125 2.51 2.50 
14 100 1.99 2.00 
15 80 1.58 1.60 
i 16 60 1.25 T720 
17 50 .99 1.00 
18 4o .80 .80 


Wiring for Motor Circuits. 


Motors are used for operating machinery of various de- 
scriptions. No large manufactory is planned without giving 
consideration to the use of electric power and wherever the 
electric current is available the progressive shop, no matter 
how small, is sure to install a motor to drive the machinery. 

To obtain good regulation is one of the features necessary 
for a motor installation: Too great a drop in the wire will 
give a voltage too low when the motor has a full load, and 
the power will be less than needed for the work when the 
greatest power is desired. 

A correct distribution of the circuits and a correct size of 
wire adds materially to the proper operation of, the motors. 

The following tables were compiled for the use of motor- 
men in their work in Chicago and the first issue consisted 
of 100 blue prints. In 1892 they were published in pam- 
phlet form. 

These tables have been carefully revised, and a new table 
on large sized feeders with ground return has been added. 


Wiring for Motor Circuits. 


FOR DIRECT CURRENT MOTORS. 


The size of a motor, which is given in horse power, is the 
maximum power that may be safely developed at the pulley. 
To obtain this mechanical power from the electrical energy 
there is a loss, and consequently there is more than one horse 
power of electrical energy supplied to a motor for every 
mechanical horse power developed at the pulley, 

To determine the SIZE OF WIRE FOR A MOTOR CIRCUIT, it is 
necessary, first, toknowthe number of amperes that will develcp 
the horse power. The electrical energy is the product of the 
voltage and the amperes.’ Therefore, the number of amperes 
required to develop one horse power will depend upon the 
voltage. In determining the number of amperes to develop a 
horse power, allowance must be made for the loss of electrical 
energy. Motors, though of the same type, yet of different 
sizes, will vary in efficiency, and motors of different types 
transform electrical energy into mechanical energy with differ- 
ent degrees of economy. To give the number of amperes 
required to develop the rated horse power on different sized 
motors, and for motors at different voltages, the table 
‘“amperes per motor’’ has been compiled, the efficiency of 
the various sized motors being taken at such a per cent that 
will approximate nearest to the actual conditions. 


sl 


AMPERES PER MOTOR. 


This table is arranged to show the amperes per motor of the 
direct current constant potential motors at the efficiencies indi- 
cated for various horse powers up to 150, aud various voltages 
up to I,200. One column shows the watts per motor, and 
another shows the number of 16-candle power lamps that equal 
the energy the motor draws from the circuit. 


A uniform loss or drop of electrical pressure in service lines 
should be established in every central station supplying cur- 
rent for power purposes. 


The pressure supplied at the motor brushes should be I1o 
volts for a 110 volt motor, and for motors of other voltages the 
pressure should be the same as the voltage at which they are 
rated. This is a point sometimes overlooked; the pressure 
supplied at the motor brushes being several volts lower than 
the pressure for which they are rated. 


Example.—What number of amperes are required to develop 
100. H. P. at 110 volts? 


Answer.—753 amperes. Follow the column marked H. P. 
down until 100 is reached and then across until column IIo 
under volts is reached; this gives 753 and is the result in 
amperes. ; 


Example.—What number of amperes are required to develop 
Sse Peat 220NV Gls. 

Answer.—I2.7 amperes. 

Example.—What number of amperes are required to develop 
50 H. P. at 500 volts? 

Answer.—82.8 amperes. 

By carefully examining the table the ease with which these 
results are obtained can be readily appreciated. 


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MINIMUM SIZE WIRK FOR MOTOR SERVICES. 


A copper conductor will not carry with safety more than a 
certain number of amperes. In the installation of a motor, 
the service wires should always be of sufficient size to carry 
the number of amperes that will be required to develop the 
maximum rated horse powerof the motor. 

It is not advisable in motor circuits to use wire smaller than 
No. 14 B. & S. gauge, as wire of smaller size is liable to be 
broken. 

The table gives the minimum size wire that should be used 
for motor services, using the table of safe carrying capacity for 
concealed wires as given by the National Board of Fire Under- 
writers, for limiting the size of service wires. 

Only the three standard voltages are given in the table, 
namely, I10, 220 and 500 volts. This table has no reference to 
the loss in voltage in the wires, but simply gives the smallest 
size wire that should at any time be used, because by using a 
smaller wire it would become heated beyond a point consid- 


ered safe by insurance companies. 


84 


MINIMUM SIZE WIRE FOR MOTOR SERVICES.— 
WHEN CONCEALED OR PARTLY CONCEALED WIRES ARE USED. 











S1zE WIRE B. AND S. GAUGE. 











| 110 Volts. 220 Volts. | 500 Volts. 
ee | . 14 Sea Ter Tiers. TA 
14 Tf | 14 
12 | 14 14 
3 | 10 | 14 | 14 
4 8 | 2 14 
5 | 6 | 10 14 
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WIRING FOR MOTOR SERVICES OR CIRCUIT'S. 


In connecting a motor the size wire which conducts the cur- 
rent from the street mains to the motor plays an important 
part in the proper running of the motor, for if it is too small 
the loss in volts will be large enough to interfere with the 
operation of the motor as the voltage at the brushes will be 
less than the motor is designed for. 

The table, ‘‘ Wiring for motor services,’’ is calculated so 
that the size wire for any loss in volts may be determined. 
The first three columns are for the horse power of the motor ; 
the fourth column, the ampere capacity required by the motor 
to develop its rated horse power. The amperes in this table 
are a close approximation to actual practice, and so long as the 
question of efficiency of the motor is a variable one, the most 
valuable electric tables must be based upon approximation de- 
rived from actual practice. In the other columns of the table 
is given the distance which the various amounts of power can 
be transmitted on different sized wires WITH A LOSS OF 
ONE VOLT. The five lines at the top of the table give infor- 
mation in regard to the wires of different sizes. The safe car- 
rying capacity in this table is that which has been adopted by 
National Board of Underwriters. The method used in apply- 
ing this table is to DIVIDE THE TOTAL DISTANCE FROM THE 
STREET CONNECTIONS to the motor by the NUMBER OF VOLTS 
which are to be allowed for Loss in the service wire. THIS 
WILL GIVE THE DISTANCE FOR ONE VOLT 1,0Ss. See in what 
columns opposite the horse power this distance or the nearest 
amount to it is, and this will indicate the size wire required. 


86 

























































































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To clearly explain the workings of the table, ‘‘ Wiring for 
motor services,’’ several examples will be given illustrating 
different conditions. 


Examiple.—A 15 horse power motor, of 220 volts, is located 
4oo feet from the street connection, and eight volts is the 
amount allowed for loss in service wire ; what size service wire 
will be required ? 


Answer.—Hight volts loss in 400 feet Cistance means a loss 
of one volt for every one-eighth of 4oo feet, or one volt loss in 
50 feet. Referring to the table, we find 15 horse power under 
220 volts can be transmitted 43 feet on No. 3 wire, or 55 feet on 
No. 2 wire. No. 2 wire is the size of wire desired, as No. 3 
falls short of the requirements. 


Example.—A 110 volt five-horse power motor is located 380 
feet from street connection and Io volts loss to be allowed? 


Answer.—Dividing 380 feet by 10 gives 38 feet for one volt. 
Referring to the table we find 38 feet, opposite five-horse 
power Ito volt motor, under No. 5 wire. 


Example.—A 30-horse power 500 volt motor is located 600 
feet from street connection, 10 volts loss allowed? 


Answer.—Six hundred feet divided by Io gives 60 feet for 
one volt. Sixty-four feet is the nearest to 60 feet, opposite 30- 
horse power motor, and indicates No. 2 wire. 


88 


A Table for Large Size Feeders with Ground Returns. 


A street railway circuit consists primarily of the overhead 
trolley wire and the return through the ground. 

The return through the ground is supplemented by the cir- 
cuit offered by the conductivity of the iron rails which is 
further increased by bonding them at the joints. The iron 
rails may also be connected at different points with copper 
conductors which lead direct to the source of supply. 


The overhead trolley wire may be the sole means of convey- 
ing the current from the power house, but it also may be sup- 
plemented by other copper conductors or feeders which carry 
the current to central points for more even distribution. 


To estimate the size of these feeders the same problem is to be 
solved that is met in determining the size of wire for ordinary 
motor circuits, with this indifference. In an ordinary motor 
circuit both the outgoing and return circuit are of copper and 
the resistance of the outgoing and return circuits are approxi- 
mately the same, and if one volt is lost on the outgoing cir- 
cuit for a certain predetermined load, then an equal loss is 
occasioned on the return circuit, and the total loss on the con- 
ducting wires is two volts. In the railway circuit, if the return 
circuit is through the rails and it is desired to put in a feeder 
connecting the trolley wire at any point, it is best to assume a 
certain number of volts loss for the transmission of the current 
through the feeder alone and determine its size, but always 
remembering that the loss in the feeder does not represent the 
total drop or loss in volts between the motor and the gener- 
ator. 

One volt will be required to transmit one ampere if the con- 
ductor is one foot and its areas in circular mils 10.6 (this is 
true of commercially pure copper at ordinary temperatures, 
see table ‘‘ Resistance of Copper at Various Temperatures ’’ ) 

Using the above as a basis, the accompanying table has been 
compiled, showing the distance at which different sized cables 


oy) 


will conduct various loads in atnperes with one volt loss, The 
first step in using the table is to ascertain the distance the 
cable extends; divide this distance by the volts it is desired to 
lose, follow this horizontal line to the figures (larger rather 
than smaller amount) nearest to the amount found by dividing 
the length of the cable by the volts. In the vertical column 
at the top is indicated the size of cable used. 


In using this table it must be remembered that no account 
is taken of the additional volts loss which exists in the return 
circuit, and to lessen those the only way is to decrease the re- 
sistance of your ground return as much as circumstances and 
capital will allow. 





Current | THE NUMBER OF FEET the different size cables will carry 





















































in tne different amount of current in amperes with 1 volt loss. 
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Amperes |g o|y 8/2 3% 8 3A e He. an om om om om a 
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50 200} 252] 316) 400 472| 556] 754! 944] 1132) 1320] 1408] 1698] 138 
100 100] 126] 158] 2-0! 23 282} 376] 472} 566, 660, 754] 848] 044 
150 . -| 84] 106) 132] 156] 188] 250] 314] 376; 440] 500] 566] 623 
200 . .{ 62] 80] 100} 118} 140} 188} 236] 282] 330! 376] 424] 472 
250 Ser [eared ite tee noe 94] 112] 150] 188] 226) 264| 300] 338] 378 
300 | eee Woncrs ee 94, 124] 156} 188! 22@! 250) 282] 314 
35? ae a | wee Peo | es Sec] ce aif = LOO) 91.3450 69) a GS eee oso 
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450 ww fw we fees SES ol =] a | Od 1241 Sr AG ie TGOlr os meee 
500 Niecy § Resin teeta Pee ellie pele? oh act Pw ac 94| I12| 132] I50] 168] 188 
600 Fe HN Peat | Pe (eral ey alfa mae IE rs Bn | re 94] Ito} 124] x40] 156 
700 ee Wee Verde is eel gee sees tlie niall ace ton Six 94} Ic6] 120] 131 
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N.B. This table takes into consideration the drop along a conductor 
from one point to another and differs in this respect from most tables 
which take into consideration the drop of the return circuit. 


90 


Standard Diagrams for Uniformity in Electrical 
Engineering and Patent Office 
Drawings, 


In order to standardize the various diagrams used by eles: 
trical engineers, and in Patent Office work, the Chicago 
Electrical Association appointed a committee to investigate 
the matter. 

The committee was appointed in January, 1807. The dia- 
grams submitted by this committee were both approved 
and adopted by the Chicago Electrical Association on Feb- 
ruary 4, 1898, and have been submitted to many engineers, 
to the experts in the Patent Office and have been discussed 
at open meetings of the Chicago Electrical Association. 
This is the first attempt of any organized body to present 
a uniform code of symbols or diagrams to the engineering 
profession. 

The list of diagrams is not complete and from time to time 
others will be added. ‘The advantage of a uniform code of 
diagrams is apparent, anda set of plans in which they are 
used would be intelligible to any engineer. 



















MOTE WME AMY OF TMESE QiacenMs Se } { 
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IMSFRT THE LETTER Gin TME OILACRAM 


~ WHEN VUSEO TO REPRESENT & MOTOR IM. 
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( ) Dynamo or Motor 
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fae 2 
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Dernres Wound 
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Shunt Wound 


Dynamo oOo Motor 


Comyrorumnd “Worn : 
Dynamo Or Motor 





On Sass 
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Sui Shave 
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Chicago Sectrical Closcciatven. 


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Magneto Sencrator 


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